[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:03.12,Default,,0000,0000,0000,,Today we'll be covering\Nsection 12.4 on the addition
Dialogue: 0,0:00:03.12,0:00:05.04,Default,,0000,0000,0000,,and complement rules.
Dialogue: 0,0:00:05.04,0:00:07.98,Default,,0000,0000,0000,,Before we begin, I\Nwant to say that most
Dialogue: 0,0:00:07.98,0:00:12.03,Default,,0000,0000,0000,,of the problems in this section\Ncan be solved using techniques
Dialogue: 0,0:00:12.03,0:00:13.47,Default,,0000,0000,0000,,we already know.
Dialogue: 0,0:00:13.47,0:00:15.99,Default,,0000,0000,0000,,But occasionally, we'll\Ncome across certain problems
Dialogue: 0,0:00:15.99,0:00:19.17,Default,,0000,0000,0000,,when we need to learn a little\Nbit more about probability.
Dialogue: 0,0:00:19.17,0:00:21.06,Default,,0000,0000,0000,,And occasionally, we'll\Nfind out that it's
Dialogue: 0,0:00:21.06,0:00:23.67,Default,,0000,0000,0000,,convenient to use\Nsome more formulas
Dialogue: 0,0:00:23.67,0:00:25.72,Default,,0000,0000,0000,,and have them in our tool box.
Dialogue: 0,0:00:25.72,0:00:28.50,Default,,0000,0000,0000,,So I'm going to do something\Nthat I don't usually do.
Dialogue: 0,0:00:28.50,0:00:32.10,Default,,0000,0000,0000,,I'm going to go ahead and give\Nyou a listing of the formulas
Dialogue: 0,0:00:32.10,0:00:33.99,Default,,0000,0000,0000,,that we're going to\Nadd to our tool box
Dialogue: 0,0:00:33.99,0:00:36.31,Default,,0000,0000,0000,,in advance of when we've\Nactually talked about them.
Dialogue: 0,0:00:36.31,0:00:37.77,Default,,0000,0000,0000,,I think it would\Nbe good to let you
Dialogue: 0,0:00:37.77,0:00:40.26,Default,,0000,0000,0000,,look at what we're going\Nto be using, mainly
Dialogue: 0,0:00:40.26,0:00:41.66,Default,,0000,0000,0000,,for the reason\Nthat said earlier.
Dialogue: 0,0:00:41.66,0:00:43.89,Default,,0000,0000,0000,,We don't always even need them.
Dialogue: 0,0:00:43.89,0:00:46.09,Default,,0000,0000,0000,,But they will be\Nhere we do need them.
Dialogue: 0,0:00:46.09,0:00:48.48,Default,,0000,0000,0000,,The first is called\Nthe addition or union
Dialogue: 0,0:00:48.48,0:00:50.13,Default,,0000,0000,0000,,rule for probabilities.
Dialogue: 0,0:00:50.13,0:00:53.25,Default,,0000,0000,0000,,It simply says that the\Nprobability of E or F
Dialogue: 0,0:00:53.25,0:00:57.45,Default,,0000,0000,0000,,is equal to the probability\Nof E plus the probability of F
Dialogue: 0,0:00:57.45,0:01:02.42,Default,,0000,0000,0000,,minus the probability of E and\NF. And remember, or means union
Dialogue: 0,0:01:02.42,0:01:04.74,Default,,0000,0000,0000,,and and means intersection.
Dialogue: 0,0:01:04.74,0:01:07.87,Default,,0000,0000,0000,,The complement rule--\Nif E is an event,
Dialogue: 0,0:01:07.87,0:01:11.65,Default,,0000,0000,0000,,and E with a superscript c is\Nthe complement of the event,
Dialogue: 0,0:01:11.65,0:01:14.69,Default,,0000,0000,0000,,then the probability of E\Ncompliment is equal to 1 minus
Dialogue: 0,0:01:14.69,0:01:18.29,Default,,0000,0000,0000,,the probability of E. The\Nproduct rule for independent
Dialogue: 0,0:01:18.29,0:01:20.15,Default,,0000,0000,0000,,and dependent events--
Dialogue: 0,0:01:20.15,0:01:23.40,Default,,0000,0000,0000,,if two events E and\NF are independent,
Dialogue: 0,0:01:23.40,0:01:26.81,Default,,0000,0000,0000,,then the probability of E and F\Nis simply the probability of E
Dialogue: 0,0:01:26.81,0:01:31.97,Default,,0000,0000,0000,,times the probability of F.\NNow the dependent version
Dialogue: 0,0:01:31.97,0:01:37.58,Default,,0000,0000,0000,,of this rule uses a notation\Nwe haven't even introduced yet,
Dialogue: 0,0:01:37.58,0:01:39.89,Default,,0000,0000,0000,,but I'm going to go\Nahead and do it anyway.
Dialogue: 0,0:01:39.89,0:01:42.11,Default,,0000,0000,0000,,We'll come back to it\Nlater and talk more
Dialogue: 0,0:01:42.11,0:01:44.29,Default,,0000,0000,0000,,about what the new\Nnotation means.
Dialogue: 0,0:01:44.29,0:01:47.78,Default,,0000,0000,0000,,But it says, if two\Nevents in F are dependent,
Dialogue: 0,0:01:47.78,0:01:50.21,Default,,0000,0000,0000,,then the probability\Nof E and F is
Dialogue: 0,0:01:50.21,0:01:53.21,Default,,0000,0000,0000,,equal to the\Nprobability of E times
Dialogue: 0,0:01:53.21,0:01:57.08,Default,,0000,0000,0000,,the probability of F\Ngiven E, which is also
Dialogue: 0,0:01:57.08,0:01:59.93,Default,,0000,0000,0000,,equal to the\Nprobability of F times
Dialogue: 0,0:01:59.93,0:02:02.69,Default,,0000,0000,0000,,the probability of\NE given F. Remember,
Dialogue: 0,0:02:02.69,0:02:05.54,Default,,0000,0000,0000,,I haven't even\Nintroduced that notation
Dialogue: 0,0:02:05.54,0:02:09.08,Default,,0000,0000,0000,,whether the vertical line is\Nused here in this context yet,
Dialogue: 0,0:02:09.08,0:02:10.31,Default,,0000,0000,0000,,but I will.
Dialogue: 0,0:02:10.31,0:02:13.67,Default,,0000,0000,0000,,That leads to something called\Nconditional probability.
Dialogue: 0,0:02:13.67,0:02:16.01,Default,,0000,0000,0000,,And I've got two\Nformulas related to that.
Dialogue: 0,0:02:16.01,0:02:20.03,Default,,0000,0000,0000,,One says the\Nprobability of E given F
Dialogue: 0,0:02:20.03,0:02:23.39,Default,,0000,0000,0000,,is equal to the\Nprobability of E and F
Dialogue: 0,0:02:23.39,0:02:27.47,Default,,0000,0000,0000,,divided by the probability of\NF. Remember when I say E and F,
Dialogue: 0,0:02:27.47,0:02:30.17,Default,,0000,0000,0000,,and means intersection.
Dialogue: 0,0:02:30.17,0:02:32.93,Default,,0000,0000,0000,,Also, the probability\Nof F given E
Dialogue: 0,0:02:32.93,0:02:35.93,Default,,0000,0000,0000,,is the probability\Nof E and F divided
Dialogue: 0,0:02:35.93,0:02:39.04,Default,,0000,0000,0000,,by the probability of E.
Dialogue: 0,0:02:39.04,0:02:41.38,Default,,0000,0000,0000,,Before I leave this,\NI told you I'm going
Dialogue: 0,0:02:41.38,0:02:43.09,Default,,0000,0000,0000,,to rotate back to this later.
Dialogue: 0,0:02:43.09,0:02:46.66,Default,,0000,0000,0000,,But I will say if\Nyou use that notation
Dialogue: 0,0:02:46.66,0:02:48.82,Default,,0000,0000,0000,,with the vertical\Nline, it's read
Dialogue: 0,0:02:48.82,0:02:53.56,Default,,0000,0000,0000,,the probability of E given\NF. The vertical line is
Dialogue: 0,0:02:53.56,0:02:56.89,Default,,0000,0000,0000,,read as given or given that.
Dialogue: 0,0:02:56.89,0:03:03.01,Default,,0000,0000,0000,,So of course, the probability of\NF given E when the E and the F
Dialogue: 0,0:03:03.01,0:03:05.98,Default,,0000,0000,0000,,are flipped around\Nthe vertical line.
Dialogue: 0,0:03:05.98,0:03:08.30,Default,,0000,0000,0000,,Like I said, we'll come\Nback to this later.
Dialogue: 0,0:03:08.30,0:03:12.13,Default,,0000,0000,0000,,But this is the addition to\Nour tool box of formulas.
Dialogue: 0,0:03:12.13,0:03:14.32,Default,,0000,0000,0000,,Sometimes we'll need them.
Dialogue: 0,0:03:14.32,0:03:16.99,Default,,0000,0000,0000,,Sometimes we'll just be\Nable to go back and do
Dialogue: 0,0:03:16.99,0:03:21.41,Default,,0000,0000,0000,,these problems using the\Nthings we already had.
Dialogue: 0,0:03:21.41,0:03:24.29,Default,,0000,0000,0000,,But they're there\Nif you need them.
Dialogue: 0,0:03:24.29,0:03:27.17,Default,,0000,0000,0000,,Suppose we want to calculate the\Nprobability of rolling two dice
Dialogue: 0,0:03:27.17,0:03:30.05,Default,,0000,0000,0000,,and getting a sum\Nof either 5 or 9.
Dialogue: 0,0:03:30.05,0:03:31.82,Default,,0000,0000,0000,,And you notice, I\Ncall this getting
Dialogue: 0,0:03:31.82,0:03:34.97,Default,,0000,0000,0000,,a sum of 5 an event\Ncalled E sub 1,
Dialogue: 0,0:03:34.97,0:03:38.54,Default,,0000,0000,0000,,and the sum of getting a\N9 an event called E sub 2.
Dialogue: 0,0:03:38.54,0:03:40.62,Default,,0000,0000,0000,,That's just for convenience.
Dialogue: 0,0:03:40.62,0:03:42.50,Default,,0000,0000,0000,,We're trying to roll\Ntwo dice and find
Dialogue: 0,0:03:42.50,0:03:46.18,Default,,0000,0000,0000,,the probability of getting\Na sum of either 5 or 9.
Dialogue: 0,0:03:46.18,0:03:49.07,Default,,0000,0000,0000,,You really don't need any of\Nthe new formulas to do this.
Dialogue: 0,0:03:49.07,0:03:50.74,Default,,0000,0000,0000,,This is just like\Na problem we might
Dialogue: 0,0:03:50.74,0:03:52.30,Default,,0000,0000,0000,,have been able to work earlier.
Dialogue: 0,0:03:52.30,0:03:53.83,Default,,0000,0000,0000,,And I want to emphasize that.
Dialogue: 0,0:03:53.83,0:03:57.13,Default,,0000,0000,0000,,What you really need to do\Nis build your 2 by 2 table
Dialogue: 0,0:03:57.13,0:04:00.79,Default,,0000,0000,0000,,for rolling two dice, which\Ngives you the sample space.
Dialogue: 0,0:04:00.79,0:04:01.81,Default,,0000,0000,0000,,And then look.
Dialogue: 0,0:04:01.81,0:04:03.25,Default,,0000,0000,0000,,We're looking for a sum of 5.
Dialogue: 0,0:04:03.25,0:04:04.91,Default,,0000,0000,0000,,That's the event I call E sub 1.
Dialogue: 0,0:04:04.91,0:04:06.12,Default,,0000,0000,0000,,There are four possibilities.
Dialogue: 0,0:04:06.12,0:04:12.93,Default,,0000,0000,0000,,You roll 4, 1; 3, 2; 2,\N3; or 1, 4; or sum of 9.
Dialogue: 0,0:04:12.93,0:04:14.67,Default,,0000,0000,0000,,That's the event\NI called E sub 2.
Dialogue: 0,0:04:14.67,0:04:17.89,Default,,0000,0000,0000,,It happened to be exactly\Nfour possibilities for that
Dialogue: 0,0:04:17.89,0:04:22.69,Default,,0000,0000,0000,,as well, either a 6, 3;\Na 5, 4, a 4, 5; or 3, 6.
Dialogue: 0,0:04:22.69,0:04:26.17,Default,,0000,0000,0000,,So the probability of\Nthe sum is either 5 or 9
Dialogue: 0,0:04:26.17,0:04:30.14,Default,,0000,0000,0000,,is simply how many ways\Nyou can get a 5 or a 9,
Dialogue: 0,0:04:30.14,0:04:34.51,Default,,0000,0000,0000,,which happens to be 8,\Nover the number of ways
Dialogue: 0,0:04:34.51,0:04:38.02,Default,,0000,0000,0000,,you could get any sum, which as\Nwe know from earlier problems
Dialogue: 0,0:04:38.02,0:04:39.78,Default,,0000,0000,0000,,is 36.
Dialogue: 0,0:04:39.78,0:04:42.67,Default,,0000,0000,0000,,Web Assign will accept\N836 as the answer.
Dialogue: 0,0:04:42.67,0:04:44.44,Default,,0000,0000,0000,,Or you can reduce it to 2/9.
Dialogue: 0,0:04:44.44,0:04:48.52,Default,,0000,0000,0000,,
Dialogue: 0,0:04:48.52,0:04:51.37,Default,,0000,0000,0000,,But there is something about\Nthose last two examples
Dialogue: 0,0:04:51.37,0:04:52.12,Default,,0000,0000,0000,,worth noting.
Dialogue: 0,0:04:52.12,0:04:54.45,Default,,0000,0000,0000,,There's a difference.
Dialogue: 0,0:04:54.45,0:04:55.29,Default,,0000,0000,0000,,Look at those two.
Dialogue: 0,0:04:55.29,0:04:56.53,Default,,0000,0000,0000,,I'm flashing on the left.
Dialogue: 0,0:04:56.53,0:04:57.69,Default,,0000,0000,0000,,The first one I did.
Dialogue: 0,0:04:57.69,0:05:00.49,Default,,0000,0000,0000,,On the right, the\Nsecond one I did.
Dialogue: 0,0:05:00.49,0:05:02.44,Default,,0000,0000,0000,,Going back to the\None on the left,
Dialogue: 0,0:05:02.44,0:05:05.87,Default,,0000,0000,0000,,the probability of\Nthe sum was 5 or 7.
Dialogue: 0,0:05:05.87,0:05:10.00,Default,,0000,0000,0000,,It turned out just to be\Nthe sum of the probability
Dialogue: 0,0:05:10.00,0:05:13.93,Default,,0000,0000,0000,,that the sum was 5 and the\Nprobability of the sum was 7.
Dialogue: 0,0:05:13.93,0:05:20.35,Default,,0000,0000,0000,,Other words, there were four\Nevents that gave me a sum of 5.
Dialogue: 0,0:05:20.35,0:05:24.96,Default,,0000,0000,0000,,So the probability of\Ngetting a sum of 5 is 436.
Dialogue: 0,0:05:24.96,0:05:28.29,Default,,0000,0000,0000,,But there are also four\Nthat gave me a sum of 8.
Dialogue: 0,0:05:28.29,0:05:33.03,Default,,0000,0000,0000,,So the probability of getting\Na sum of 8 is also 436.
Dialogue: 0,0:05:33.03,0:05:37.94,Default,,0000,0000,0000,,And 436 plus 436 is 836.
Dialogue: 0,0:05:37.94,0:05:41.54,Default,,0000,0000,0000,,But if you look\Non the right side,
Dialogue: 0,0:05:41.54,0:05:44.15,Default,,0000,0000,0000,,the probability that a\Nnumber is greater than 2
Dialogue: 0,0:05:44.15,0:05:47.50,Default,,0000,0000,0000,,or odd turned out to not\Nbe equal to the probability
Dialogue: 0,0:05:47.50,0:05:51.34,Default,,0000,0000,0000,,that the sum was 2 plus the\Nprobability of the sum was odd.
Dialogue: 0,0:05:51.34,0:05:53.21,Default,,0000,0000,0000,,If you look at it, the\Nprobability of the sum
Dialogue: 0,0:05:53.21,0:05:58.61,Default,,0000,0000,0000,,was 2, 3, 4, 5, 6, 7, 8,\N9, 10 was 8 out of 10.
Dialogue: 0,0:05:58.61,0:06:02.21,Default,,0000,0000,0000,,The probability of the sum\Nwas odd was 5 out of 10.
Dialogue: 0,0:06:02.21,0:06:07.53,Default,,0000,0000,0000,,Well obviously, 8/10 plus\N5/10 is 13/10 not 9/10.
Dialogue: 0,0:06:07.53,0:06:10.70,Default,,0000,0000,0000,,So on the left, I was just\Nable to take the two events
Dialogue: 0,0:06:10.70,0:06:12.68,Default,,0000,0000,0000,,separately and add\Nup the probabilities
Dialogue: 0,0:06:12.68,0:06:14.42,Default,,0000,0000,0000,,and get the correct answer.
Dialogue: 0,0:06:14.42,0:06:20.47,Default,,0000,0000,0000,,On the right, if I tried\Nthat trick it wouldn't work.
Dialogue: 0,0:06:20.47,0:06:21.76,Default,,0000,0000,0000,,Why is that?
Dialogue: 0,0:06:21.76,0:06:23.55,Default,,0000,0000,0000,,It's an important distinction.
Dialogue: 0,0:06:23.55,0:06:24.76,Default,,0000,0000,0000,,Think about it just a minute.
Dialogue: 0,0:06:24.76,0:06:27.80,Default,,0000,0000,0000,,
Dialogue: 0,0:06:27.80,0:06:30.74,Default,,0000,0000,0000,,In a nutshell, it's because on\Nthe right some of the values
Dialogue: 0,0:06:30.74,0:06:32.75,Default,,0000,0000,0000,,are greater than 2 and odd.
Dialogue: 0,0:06:32.75,0:06:36.14,Default,,0000,0000,0000,,Remember that when we did\Nthis problem on the right,
Dialogue: 0,0:06:36.14,0:06:38.57,Default,,0000,0000,0000,,there were certain numbers\Nthat were both greater than 2
Dialogue: 0,0:06:38.57,0:06:39.11,Default,,0000,0000,0000,,and odd.
Dialogue: 0,0:06:39.11,0:06:41.84,Default,,0000,0000,0000,,For example, 3 is\Ngreater than 2 and odd.
Dialogue: 0,0:06:41.84,0:06:43.97,Default,,0000,0000,0000,,5 is greater than 2 and odd.
Dialogue: 0,0:06:43.97,0:06:46.73,Default,,0000,0000,0000,,7 is greater than 2 and odd.
Dialogue: 0,0:06:46.73,0:06:49.91,Default,,0000,0000,0000,,And 9 is greater than 2 and odd.
Dialogue: 0,0:06:49.91,0:06:52.79,Default,,0000,0000,0000,,So there's an overlap\Nbetween those counts.
Dialogue: 0,0:06:52.79,0:06:54.77,Default,,0000,0000,0000,,On the left, that didn't happen.
Dialogue: 0,0:06:54.77,0:06:57.86,Default,,0000,0000,0000,,When you looked at the\Nnumbers that summed to 5
Dialogue: 0,0:06:57.86,0:07:01.53,Default,,0000,0000,0000,,and the numbers that summed\Nto 7, there was no overlap.
Dialogue: 0,0:07:01.53,0:07:05.67,Default,,0000,0000,0000,,That's what caused\Nthe difference.
Dialogue: 0,0:07:05.67,0:07:08.16,Default,,0000,0000,0000,,Noting this leads to\Nan important result
Dialogue: 0,0:07:08.16,0:07:10.84,Default,,0000,0000,0000,,called the addition or union\Nrule for probabilities.
Dialogue: 0,0:07:10.84,0:07:13.65,Default,,0000,0000,0000,,Remember, that's one of the\Nones that I flashed up earlier.
Dialogue: 0,0:07:13.65,0:07:16.62,Default,,0000,0000,0000,,It says that the\Nprobability of E or F
Dialogue: 0,0:07:16.62,0:07:20.52,Default,,0000,0000,0000,,is equal the probability of\NE plus the probability of F
Dialogue: 0,0:07:20.52,0:07:23.07,Default,,0000,0000,0000,,minus the probability\Nof E and F.
Dialogue: 0,0:07:23.07,0:07:26.79,Default,,0000,0000,0000,,And if you look at that formula,\Nit encapsulates the difference
Dialogue: 0,0:07:26.79,0:07:31.30,Default,,0000,0000,0000,,that we just noticed between\Nthose two early examples.
Dialogue: 0,0:07:31.30,0:07:33.55,Default,,0000,0000,0000,,You could have used\Nthis rule to solve
Dialogue: 0,0:07:33.55,0:07:36.82,Default,,0000,0000,0000,,both of the earlier problems,\Nbut you don't have to.
Dialogue: 0,0:07:36.82,0:07:38.53,Default,,0000,0000,0000,,We were able to figure\Nout a way to do it
Dialogue: 0,0:07:38.53,0:07:41.10,Default,,0000,0000,0000,,without using the formula.
Dialogue: 0,0:07:41.10,0:07:43.81,Default,,0000,0000,0000,,But I will say that if\Nyou did use the formula
Dialogue: 0,0:07:43.81,0:07:46.42,Default,,0000,0000,0000,,for the first one,\Nwhere the advance were
Dialogue: 0,0:07:46.42,0:07:47.81,Default,,0000,0000,0000,,mutually exclusive--
Dialogue: 0,0:07:47.81,0:07:51.85,Default,,0000,0000,0000,,the sum being 5, there was no\Noverlap with the sum being 7.
Dialogue: 0,0:07:51.85,0:07:53.44,Default,,0000,0000,0000,,There was nothing\Nto subtract away.
Dialogue: 0,0:07:53.44,0:07:55.87,Default,,0000,0000,0000,,If you take that minus\Npart of the formula
Dialogue: 0,0:07:55.87,0:07:57.64,Default,,0000,0000,0000,,it would have been 0.
Dialogue: 0,0:07:57.64,0:08:01.75,Default,,0000,0000,0000,,The probability would be 0\Nthat a sum could be 5 and 7.
Dialogue: 0,0:08:01.75,0:08:05.56,Default,,0000,0000,0000,,However, on the second\None, there was an overlap.
Dialogue: 0,0:08:05.56,0:08:09.46,Default,,0000,0000,0000,,There were numbers bigger\Nthan 2 that were also odd.
Dialogue: 0,0:08:09.46,0:08:13.05,Default,,0000,0000,0000,,This formula takes care of that.
Dialogue: 0,0:08:13.05,0:08:14.94,Default,,0000,0000,0000,,Think about that\Nwhat that means.
Dialogue: 0,0:08:14.94,0:08:17.32,Default,,0000,0000,0000,,If the events are\Nmutually exclusive,
Dialogue: 0,0:08:17.32,0:08:20.49,Default,,0000,0000,0000,,the probability of them both\Noccurring together is 0.
Dialogue: 0,0:08:20.49,0:08:23.08,Default,,0000,0000,0000,,So you're actually\Nsubtracting nothing.
Dialogue: 0,0:08:23.08,0:08:25.92,Default,,0000,0000,0000,,So the formula also\Nallows for the possibility
Dialogue: 0,0:08:25.92,0:08:29.42,Default,,0000,0000,0000,,that that overlap is\Nnot probability 0.
Dialogue: 0,0:08:29.42,0:08:30.75,Default,,0000,0000,0000,,So the formula works either way.
Dialogue: 0,0:08:30.75,0:08:32.19,Default,,0000,0000,0000,,But if they're\Nmutually exclusive,
Dialogue: 0,0:08:32.19,0:08:34.26,Default,,0000,0000,0000,,remember, just nothing subtract.
Dialogue: 0,0:08:34.26,0:08:37.56,Default,,0000,0000,0000,,The formula just degenerates\Nto having to subtract nothing.
Dialogue: 0,0:08:37.56,0:08:40.45,Default,,0000,0000,0000,,Now let's look at\Nsome problems where
Dialogue: 0,0:08:40.45,0:08:44.34,Default,,0000,0000,0000,,we might want to think about it\Nin terms of using this formula.
Dialogue: 0,0:08:44.34,0:08:48.69,Default,,0000,0000,0000,,If we select a single card\Nfrom a standard 52-card deck,
Dialogue: 0,0:08:48.69,0:08:51.87,Default,,0000,0000,0000,,what is the probability that we\Ndraw either a heart or a face
Dialogue: 0,0:08:51.87,0:08:52.86,Default,,0000,0000,0000,,card.
Dialogue: 0,0:08:52.86,0:08:57.37,Default,,0000,0000,0000,,So the probability of\Na heart or a face card.
Dialogue: 0,0:08:57.37,0:09:01.24,Default,,0000,0000,0000,,Recall with this standard\N52-card deck looks like.
Dialogue: 0,0:09:01.24,0:09:05.89,Default,,0000,0000,0000,,It has 4 suits, 12 face\Ncards, et cetera, et cetera.
Dialogue: 0,0:09:05.89,0:09:08.92,Default,,0000,0000,0000,,So the probability of\Na heart or a face card
Dialogue: 0,0:09:08.92,0:09:11.23,Default,,0000,0000,0000,,is just going to be\Nthe number of hearts
Dialogue: 0,0:09:11.23,0:09:15.49,Default,,0000,0000,0000,,or face cards over the\Nnumber of cards in the deck.
Dialogue: 0,0:09:15.49,0:09:17.59,Default,,0000,0000,0000,,Well, if you look at\Nthe hearts or face cards
Dialogue: 0,0:09:17.59,0:09:19.71,Default,,0000,0000,0000,,and just count\Nthem, the face cards
Dialogue: 0,0:09:19.71,0:09:21.67,Default,,0000,0000,0000,,are the cards that actually\Nhave faces on them.
Dialogue: 0,0:09:21.67,0:09:23.84,Default,,0000,0000,0000,,And you've got the hearts\Nas the other possibility.
Dialogue: 0,0:09:23.84,0:09:27.47,Default,,0000,0000,0000,,If you just count, you'll see\Nthat there are 22 of them--
Dialogue: 0,0:09:27.47,0:09:30.20,Default,,0000,0000,0000,,over the number of cards\Nin the deck, which is 52.
Dialogue: 0,0:09:30.20,0:09:32.22,Default,,0000,0000,0000,,So you get 22 over 52.
Dialogue: 0,0:09:32.22,0:09:35.96,Default,,0000,0000,0000,,And if you want to reduce\Nit, you get 11 over 26.
Dialogue: 0,0:09:35.96,0:09:39.17,Default,,0000,0000,0000,,Suppose I wanted to actually\Nuse that addition rule,
Dialogue: 0,0:09:39.17,0:09:42.02,Default,,0000,0000,0000,,rather than just figure\Nit out as I just did here.
Dialogue: 0,0:09:42.02,0:09:43.39,Default,,0000,0000,0000,,It would still work.
Dialogue: 0,0:09:43.39,0:09:45.89,Default,,0000,0000,0000,,I would just simply take the\Nprobability of getting a heart.
Dialogue: 0,0:09:45.89,0:09:48.68,Default,,0000,0000,0000,,Well remember, there are\N13 hearts in the deck.
Dialogue: 0,0:09:48.68,0:09:52.37,Default,,0000,0000,0000,,So the probability of getting\Na heart is 13 over 52.
Dialogue: 0,0:09:52.37,0:09:54.17,Default,,0000,0000,0000,,Then I would calculate\Nthe probability
Dialogue: 0,0:09:54.17,0:09:55.16,Default,,0000,0000,0000,,of getting a face card.
Dialogue: 0,0:09:55.16,0:09:57.38,Default,,0000,0000,0000,,Well, there are 12 face\Ncards in the deck--
Dialogue: 0,0:09:57.38,0:09:59.72,Default,,0000,0000,0000,,4 jacks, 4 queens, 4 kings.
Dialogue: 0,0:09:59.72,0:10:03.00,Default,,0000,0000,0000,,So the probability of getting\Na face card is 12 over 52.
Dialogue: 0,0:10:03.00,0:10:05.81,Default,,0000,0000,0000,,And if I apply the formula,\NI'll see that I still
Dialogue: 0,0:10:05.81,0:10:08.69,Default,,0000,0000,0000,,need to calculate the\Nprobability that a card is
Dialogue: 0,0:10:08.69,0:10:12.11,Default,,0000,0000,0000,,both a heart and a face card.
Dialogue: 0,0:10:12.11,0:10:14.06,Default,,0000,0000,0000,,And if you look at\Nthe picture, there
Dialogue: 0,0:10:14.06,0:10:17.41,Default,,0000,0000,0000,,are three heart that\Nare also face cards.
Dialogue: 0,0:10:17.41,0:10:19.61,Default,,0000,0000,0000,,That's the jack of hearts,\Nthe queen of hearts,
Dialogue: 0,0:10:19.61,0:10:20.82,Default,,0000,0000,0000,,and the king of hearts.
Dialogue: 0,0:10:20.82,0:10:23.96,Default,,0000,0000,0000,,So the probability that\Nthe card is both is 3/52.
Dialogue: 0,0:10:23.96,0:10:27.08,Default,,0000,0000,0000,,And if you plug in to the\Naddition rule formula,
Dialogue: 0,0:10:27.08,0:10:30.26,Default,,0000,0000,0000,,you'll notice that you get\Nthe probability of getting
Dialogue: 0,0:10:30.26,0:10:35.69,Default,,0000,0000,0000,,a heart is 13/52 plus the\Nprobability of getting a face
Dialogue: 0,0:10:35.69,0:10:39.74,Default,,0000,0000,0000,,card is 12/52 minus\Nthe probability
Dialogue: 0,0:10:39.74,0:10:42.83,Default,,0000,0000,0000,,that you get a card that's\Na face card and a heart--
Dialogue: 0,0:10:42.83,0:10:44.54,Default,,0000,0000,0000,,that's 3/52.
Dialogue: 0,0:10:44.54,0:10:49.09,Default,,0000,0000,0000,,And that's 13 plus\N12 minus 3 is 22.
Dialogue: 0,0:10:49.09,0:10:53.84,Default,,0000,0000,0000,,So you get 22 over 52, which\Nis 11/26, which is exactly
Dialogue: 0,0:10:53.84,0:10:57.83,Default,,0000,0000,0000,,the answer we got earlier\Nby doing it without using
Dialogue: 0,0:10:57.83,0:10:59.28,Default,,0000,0000,0000,,the addition rule.
Dialogue: 0,0:10:59.28,0:11:01.79,Default,,0000,0000,0000,,So this is just another\Nexample of where
Dialogue: 0,0:11:01.79,0:11:04.34,Default,,0000,0000,0000,,the addition rule works.
Dialogue: 0,0:11:04.34,0:11:06.92,Default,,0000,0000,0000,,But you could really\Ndo it without it.
Dialogue: 0,0:11:06.92,0:11:09.57,Default,,0000,0000,0000,,
Dialogue: 0,0:11:09.57,0:11:11.64,Default,,0000,0000,0000,,Here's an example\Nwhere using the rule
Dialogue: 0,0:11:11.64,0:11:14.19,Default,,0000,0000,0000,,is really the best\Nway to do it--
Dialogue: 0,0:11:14.19,0:11:15.97,Default,,0000,0000,0000,,the most straightforward\Nway to do it.
Dialogue: 0,0:11:15.97,0:11:19.47,Default,,0000,0000,0000,,It says if the probability of\NA is 0.3, the probability of B
Dialogue: 0,0:11:19.47,0:11:23.55,Default,,0000,0000,0000,,is 0.4, and the probability\Nof A and B is 0.2.
Dialogue: 0,0:11:23.55,0:11:25.98,Default,,0000,0000,0000,,Find the probability of A or B.
Dialogue: 0,0:11:25.98,0:11:30.00,Default,,0000,0000,0000,,Remember, and is\Nintersection, or is union.
Dialogue: 0,0:11:30.00,0:11:32.70,Default,,0000,0000,0000,,So I would say the most\Nstraightforward way
Dialogue: 0,0:11:32.70,0:11:36.48,Default,,0000,0000,0000,,to do this particular problem\Nis to use the formula.
Dialogue: 0,0:11:36.48,0:11:40.99,Default,,0000,0000,0000,,If you write the formula down,\Nreplacing E and F with and b,
Dialogue: 0,0:11:40.99,0:11:44.64,Default,,0000,0000,0000,,you get the probability of A or\NB is equal the probability of A
Dialogue: 0,0:11:44.64,0:11:49.98,Default,,0000,0000,0000,,plus the probability of B minus\Nthe probability of A and B.
Dialogue: 0,0:11:49.98,0:11:52.59,Default,,0000,0000,0000,,Well, we know the\Nprobability of A is 0.3.
Dialogue: 0,0:11:52.59,0:11:55.02,Default,,0000,0000,0000,,We know the probability\Nof B is 0.4.
Dialogue: 0,0:11:55.02,0:11:58.92,Default,,0000,0000,0000,,And we also know that the\Nprobability of A and B is 0.2.
Dialogue: 0,0:11:58.92,0:12:02.07,Default,,0000,0000,0000,,So a little bit of arithmetic\Ntells you that the final
Dialogue: 0,0:12:02.07,0:12:04.95,Default,,0000,0000,0000,,answer-- the\Nprobability of A or B--
Dialogue: 0,0:12:04.95,0:12:06.69,Default,,0000,0000,0000,,is 0.5.
Dialogue: 0,0:12:06.69,0:12:08.71,Default,,0000,0000,0000,,That's the most\Nstraightforward way to do it.
Dialogue: 0,0:12:08.71,0:12:11.90,Default,,0000,0000,0000,,
Dialogue: 0,0:12:11.90,0:12:13.27,Default,,0000,0000,0000,,Another example\Nwhere the formula
Dialogue: 0,0:12:13.27,0:12:16.10,Default,,0000,0000,0000,,is the most straightforward\Nway to do it,
Dialogue: 0,0:12:16.10,0:12:18.88,Default,,0000,0000,0000,,what if the probability of A\Nis 0.3, the probability of B
Dialogue: 0,0:12:18.88,0:12:23.06,Default,,0000,0000,0000,,is 0.4, the probability\Nof A or B is 0.5.
Dialogue: 0,0:12:23.06,0:12:26.25,Default,,0000,0000,0000,,But this time we're looking\Nfor the probability of A and B.
Dialogue: 0,0:12:26.25,0:12:29.29,Default,,0000,0000,0000,,So it's the same formula, but\Nyou know different things.
Dialogue: 0,0:12:29.29,0:12:33.31,Default,,0000,0000,0000,,Here, you know the\Nprobability of A or B is 0.5.
Dialogue: 0,0:12:33.31,0:12:35.66,Default,,0000,0000,0000,,The probability of A is 0.3.
Dialogue: 0,0:12:35.66,0:12:37.60,Default,,0000,0000,0000,,The probability of B is 0.4.
Dialogue: 0,0:12:37.60,0:12:40.86,Default,,0000,0000,0000,,And you're looking for the\Nprobability of A and B.
Dialogue: 0,0:12:40.86,0:12:42.99,Default,,0000,0000,0000,,It becomes a small\Nalgebra problem.
Dialogue: 0,0:12:42.99,0:12:43.80,Default,,0000,0000,0000,,It's easy enough.
Dialogue: 0,0:12:43.80,0:12:45.99,Default,,0000,0000,0000,,0.3 plus 0.4 is 0.7.
Dialogue: 0,0:12:45.99,0:12:48.54,Default,,0000,0000,0000,,If you move the negative\Nprobability to the left
Dialogue: 0,0:12:48.54,0:12:52.50,Default,,0000,0000,0000,,and move the 0.5 to the other\Nside, making it negative,
Dialogue: 0,0:12:52.50,0:12:56.10,Default,,0000,0000,0000,,you'll end up finding that\Nthe probability of A and B
Dialogue: 0,0:12:56.10,0:13:02.38,Default,,0000,0000,0000,,is 0.7 minus 0.5, which is 0.2.
Dialogue: 0,0:13:02.38,0:13:05.92,Default,,0000,0000,0000,,These are both examples where\Nusing the addition or union
Dialogue: 0,0:13:05.92,0:13:09.37,Default,,0000,0000,0000,,rule for probabilities is\Nthe most straightforward way
Dialogue: 0,0:13:09.37,0:13:11.08,Default,,0000,0000,0000,,to solve the problem.
Dialogue: 0,0:13:11.08,0:13:13.24,Default,,0000,0000,0000,,Use the following table\Nto find the probability
Dialogue: 0,0:13:13.24,0:13:16.09,Default,,0000,0000,0000,,that a randomly chosen member\Nof the student government board
Dialogue: 0,0:13:16.09,0:13:19.27,Default,,0000,0000,0000,,is a freshman or lives\Nin off-campus housing.
Dialogue: 0,0:13:19.27,0:13:20.92,Default,,0000,0000,0000,,They give you a table.
Dialogue: 0,0:13:20.92,0:13:23.77,Default,,0000,0000,0000,,This is another problem\Nwhere you probably can just
Dialogue: 0,0:13:23.77,0:13:25.63,Default,,0000,0000,0000,,do it by the old techniques.
Dialogue: 0,0:13:25.63,0:13:28.87,Default,,0000,0000,0000,,You want to know the probability\Nthat a person is a freshman
Dialogue: 0,0:13:28.87,0:13:31.35,Default,,0000,0000,0000,,or lives off campus.
Dialogue: 0,0:13:31.35,0:13:33.00,Default,,0000,0000,0000,,So you look at your chart.
Dialogue: 0,0:13:33.00,0:13:42.08,Default,,0000,0000,0000,,There are five freshmen\Nand there are five students
Dialogue: 0,0:13:42.08,0:13:43.79,Default,,0000,0000,0000,,who live off campus.
Dialogue: 0,0:13:43.79,0:13:47.78,Default,,0000,0000,0000,,So we want to know the\Nprobability of being a freshman
Dialogue: 0,0:13:47.78,0:13:50.18,Default,,0000,0000,0000,,or living off campus.
Dialogue: 0,0:13:50.18,0:13:52.70,Default,,0000,0000,0000,,You can't just say 5\Nplus 5, because there's
Dialogue: 0,0:13:52.70,0:13:54.03,Default,,0000,0000,0000,,one person who overlaps.
Dialogue: 0,0:13:54.03,0:13:55.94,Default,,0000,0000,0000,,There's one person\Nwho is a freshman,
Dialogue: 0,0:13:55.94,0:13:57.36,Default,,0000,0000,0000,,but also lives off campus.
Dialogue: 0,0:13:57.36,0:14:01.43,Default,,0000,0000,0000,,And when you count the number\Nof freshmen or students
Dialogue: 0,0:14:01.43,0:14:05.18,Default,,0000,0000,0000,,living off campus, you have\Nto say 4 plus 1 plus 4.
Dialogue: 0,0:14:05.18,0:14:08.00,Default,,0000,0000,0000,,That's only 9.
Dialogue: 0,0:14:08.00,0:14:09.86,Default,,0000,0000,0000,,And then when you\Ncount everybody,
Dialogue: 0,0:14:09.86,0:14:13.67,Default,,0000,0000,0000,,you get 4 plus 1 plus\N2 plus 0 plus 4 plus 2
Dialogue: 0,0:14:13.67,0:14:16.19,Default,,0000,0000,0000,,plus 1 plus 0 plus 4 plus 0, 0.
Dialogue: 0,0:14:16.19,0:14:17.42,Default,,0000,0000,0000,,You get 17.
Dialogue: 0,0:14:17.42,0:14:20.66,Default,,0000,0000,0000,,So the answer is 9/17.
Dialogue: 0,0:14:20.66,0:14:22.43,Default,,0000,0000,0000,,Again, you could use\Nthe addition rule
Dialogue: 0,0:14:22.43,0:14:23.39,Default,,0000,0000,0000,,to get the answer.
Dialogue: 0,0:14:23.39,0:14:27.97,Default,,0000,0000,0000,,But I would recommend just\Ndoing it straightforwardly.
Dialogue: 0,0:14:27.97,0:14:30.40,Default,,0000,0000,0000,,Consider the experiment\Nof rolling two dice.
Dialogue: 0,0:14:30.40,0:14:32.92,Default,,0000,0000,0000,,Let A be the event of\Nrolling a sum of 8.
Dialogue: 0,0:14:32.92,0:14:35.73,Default,,0000,0000,0000,,And let B be the event\Nof rolling a double.
Dialogue: 0,0:14:35.73,0:14:38.74,Default,,0000,0000,0000,,A double means you get the same\Nnumber on both dice, like a 2
Dialogue: 0,0:14:38.74,0:14:41.17,Default,,0000,0000,0000,,and a 2 or 5 and a 5.
Dialogue: 0,0:14:41.17,0:14:46.45,Default,,0000,0000,0000,,Find the probability of getting\Neither a sum of A or a double.
Dialogue: 0,0:14:46.45,0:14:48.79,Default,,0000,0000,0000,,Again, when you're\Nrolling two dice,
Dialogue: 0,0:14:48.79,0:14:50.96,Default,,0000,0000,0000,,you think of that as a\Ntwo-stage experiment.
Dialogue: 0,0:14:50.96,0:14:52.06,Default,,0000,0000,0000,,So you build the table.
Dialogue: 0,0:14:52.06,0:14:54.98,Default,,0000,0000,0000,,We've talked about that before.
Dialogue: 0,0:14:54.98,0:14:58.82,Default,,0000,0000,0000,,The probability of getting\Nan 8 sum or a double
Dialogue: 0,0:14:58.82,0:15:02.54,Default,,0000,0000,0000,,is going to be the number of\Nways of getting a sum of 8
Dialogue: 0,0:15:02.54,0:15:06.63,Default,,0000,0000,0000,,or a double divided by the\Nnumber of ways to get any sum.
Dialogue: 0,0:15:06.63,0:15:08.24,Default,,0000,0000,0000,,So if you look at\Nthe numerator, let's
Dialogue: 0,0:15:08.24,0:15:11.00,Default,,0000,0000,0000,,first look at the number of\Nways of getting a sum of 8.
Dialogue: 0,0:15:11.00,0:15:16.73,Default,,0000,0000,0000,,You could get a 6, 2; a 5,\N3; a 4, 4; a 3, 5; or a 2, 6.
Dialogue: 0,0:15:16.73,0:15:18.86,Default,,0000,0000,0000,,Then look at the number\Nof ways to get a double.
Dialogue: 0,0:15:18.86,0:15:24.26,Default,,0000,0000,0000,,You could get a 1, 1; a 2, 2; a\N3, 3; a 4, 4; a 5, 5 or a 6, 6.
Dialogue: 0,0:15:24.26,0:15:26.39,Default,,0000,0000,0000,,And you want to count\Nall those possibilities.
Dialogue: 0,0:15:26.39,0:15:28.25,Default,,0000,0000,0000,,Notice, there's an\Noverlap of 4, 4.
Dialogue: 0,0:15:28.25,0:15:29.75,Default,,0000,0000,0000,,So when you're\Ncounting, you don't
Dialogue: 0,0:15:29.75,0:15:31.82,Default,,0000,0000,0000,,want to count that 4, 4 twice.
Dialogue: 0,0:15:31.82,0:15:40.37,Default,,0000,0000,0000,,If you count them, you get 1,\N2, 3, 4, 5, 6, 7, 8, 9, 10.
Dialogue: 0,0:15:40.37,0:15:42.95,Default,,0000,0000,0000,,Don't count that 4, 4 twice.
Dialogue: 0,0:15:42.95,0:15:46.97,Default,,0000,0000,0000,,So that gives you 10 over the\Nnumber of ways to get any sum.
Dialogue: 0,0:15:46.97,0:15:49.61,Default,,0000,0000,0000,,Well it's, as we've learned\Nfrom doing these type
Dialogue: 0,0:15:49.61,0:15:52.88,Default,,0000,0000,0000,,problems multiple times,\Nthere are 36 possibilities.
Dialogue: 0,0:15:52.88,0:15:54.77,Default,,0000,0000,0000,,So the answer is 10 over 36.
Dialogue: 0,0:15:54.77,0:15:59.00,Default,,0000,0000,0000,,And if you want to\Nreduce it you get 5/18.
Dialogue: 0,0:15:59.00,0:16:03.53,Default,,0000,0000,0000,,Again, we didn't have to\Nexplicitly use the addition
Dialogue: 0,0:16:03.53,0:16:06.82,Default,,0000,0000,0000,,rule for probabilities,\Nalthough you could.
Dialogue: 0,0:16:06.82,0:16:08.44,Default,,0000,0000,0000,,Let's go to the\Ncomplement rule that we
Dialogue: 0,0:16:08.44,0:16:11.02,Default,,0000,0000,0000,,introduced in our\Ntoolbox at the beginning.
Dialogue: 0,0:16:11.02,0:16:12.49,Default,,0000,0000,0000,,The complement of an event.
Dialogue: 0,0:16:12.49,0:16:13.99,Default,,0000,0000,0000,,Let's take an\Nexperiment where we're
Dialogue: 0,0:16:13.99,0:16:17.02,Default,,0000,0000,0000,,rolling a single ordinary dice.
Dialogue: 0,0:16:17.02,0:16:22.65,Default,,0000,0000,0000,,The sample space obviously, is\Ngoing to be either 1, 2, 3, 4,
Dialogue: 0,0:16:22.65,0:16:24.30,Default,,0000,0000,0000,,5, or 6.
Dialogue: 0,0:16:24.30,0:16:25.76,Default,,0000,0000,0000,,So if we want to\Nlook for the event
Dialogue: 0,0:16:25.76,0:16:29.65,Default,,0000,0000,0000,,that we rolled a 3, the\Nprobability of doing
Dialogue: 0,0:16:29.65,0:16:31.96,Default,,0000,0000,0000,,that is the number\Nof ways you can
Dialogue: 0,0:16:31.96,0:16:35.02,Default,,0000,0000,0000,,get a 3, which if you're\Nrolling a single die,
Dialogue: 0,0:16:35.02,0:16:37.87,Default,,0000,0000,0000,,there's only one way to do that,\Nover the number of elements
Dialogue: 0,0:16:37.87,0:16:38.81,Default,,0000,0000,0000,,in the sample space.
Dialogue: 0,0:16:38.81,0:16:41.80,Default,,0000,0000,0000,,So the probability of\Nrolling a 3, is 1/6.
Dialogue: 0,0:16:41.80,0:16:44.31,Default,,0000,0000,0000,,
Dialogue: 0,0:16:44.31,0:16:46.63,Default,,0000,0000,0000,,The complement of that event--
Dialogue: 0,0:16:46.63,0:16:48.90,Default,,0000,0000,0000,,and we're going to\Nuse the notation
Dialogue: 0,0:16:48.90,0:16:52.40,Default,,0000,0000,0000,,E with the superscript c--
Dialogue: 0,0:16:52.40,0:16:56.22,Default,,0000,0000,0000,,some text use E with\Nan accent symbol
Dialogue: 0,0:16:56.22,0:17:00.10,Default,,0000,0000,0000,,like we used back when we talked\Nabout regular set operations.
Dialogue: 0,0:17:00.10,0:17:04.15,Default,,0000,0000,0000,,But if you see it either way,\Nit means the same thing--
Dialogue: 0,0:17:04.15,0:17:09.31,Default,,0000,0000,0000,,E with a superscript of c\Nor E with an accent symbol.
Dialogue: 0,0:17:09.31,0:17:12.89,Default,,0000,0000,0000,,It's sort of in a sense,\Nthe opposite of the event.
Dialogue: 0,0:17:12.89,0:17:16.48,Default,,0000,0000,0000,,So if you're looking\Nfor an outcome,
Dialogue: 0,0:17:16.48,0:17:21.64,Default,,0000,0000,0000,,the complement is everything\Nexcept that outcome.
Dialogue: 0,0:17:21.64,0:17:23.22,Default,,0000,0000,0000,,So if you're talking\Nabout the problem
Dialogue: 0,0:17:23.22,0:17:29.34,Default,,0000,0000,0000,,we did above, if the\Nevent E was getting a 3,
Dialogue: 0,0:17:29.34,0:17:33.12,Default,,0000,0000,0000,,then E complement\Nis not getting a 3.
Dialogue: 0,0:17:33.12,0:17:35.88,Default,,0000,0000,0000,,So in this case, if we wanted\Nto write out E complement,
Dialogue: 0,0:17:35.88,0:17:37.66,Default,,0000,0000,0000,,it would be everything except 3.
Dialogue: 0,0:17:37.66,0:17:40.90,Default,,0000,0000,0000,,So it would be 1, 2, 4, 5, or 6.
Dialogue: 0,0:17:40.90,0:17:43.23,Default,,0000,0000,0000,,So when we're talking about\Nthe complement event, that's
Dialogue: 0,0:17:43.23,0:17:48.24,Default,,0000,0000,0000,,what we mean, everything\Nbut what makes up the event.
Dialogue: 0,0:17:48.24,0:17:52.17,Default,,0000,0000,0000,,There are five things that\Naren't 3's out of possible 6.
Dialogue: 0,0:17:52.17,0:17:53.58,Default,,0000,0000,0000,,So you get 5/6.
Dialogue: 0,0:17:53.58,0:17:57.63,Default,,0000,0000,0000,,And notice that the probability\Nof getting a 3 is 1/6,
Dialogue: 0,0:17:57.63,0:18:01.83,Default,,0000,0000,0000,,and the probability of\Nnot getting a 3 is 5/6.
Dialogue: 0,0:18:01.83,0:18:05.39,Default,,0000,0000,0000,,And that's not a coincidence.
Dialogue: 0,0:18:05.39,0:18:07.64,Default,,0000,0000,0000,,That's what the\Ncomplement rule says.
Dialogue: 0,0:18:07.64,0:18:10.79,Default,,0000,0000,0000,,It simply says if you know\Nthe probability of an event,
Dialogue: 0,0:18:10.79,0:18:13.04,Default,,0000,0000,0000,,and you want to know the\Nprobability of the complement
Dialogue: 0,0:18:13.04,0:18:15.86,Default,,0000,0000,0000,,event, you just take 1\Nminus the probability of E.
Dialogue: 0,0:18:15.86,0:18:19.36,Default,,0000,0000,0000,,And importance of that rule\Nis-- it same sort of obvious--
Dialogue: 0,0:18:19.36,0:18:21.17,Default,,0000,0000,0000,,but it's very useful\Nbecause some problems,
Dialogue: 0,0:18:21.17,0:18:24.41,Default,,0000,0000,0000,,it's really easy to calculate\Nthe probability of an event,
Dialogue: 0,0:18:24.41,0:18:27.50,Default,,0000,0000,0000,,and it's a little more difficult\Nor more than a little more
Dialogue: 0,0:18:27.50,0:18:30.98,Default,,0000,0000,0000,,difficult sometimes to calculate\Ndirectly, the probability
Dialogue: 0,0:18:30.98,0:18:31.73,Default,,0000,0000,0000,,of the complement.
Dialogue: 0,0:18:31.73,0:18:34.28,Default,,0000,0000,0000,,But you can calculate\Nthe probability event,
Dialogue: 0,0:18:34.28,0:18:36.62,Default,,0000,0000,0000,,all you have to do is say\N1 minus that probability,
Dialogue: 0,0:18:36.62,0:18:38.78,Default,,0000,0000,0000,,and you'll automatically\Nhave the probability
Dialogue: 0,0:18:38.78,0:18:40.82,Default,,0000,0000,0000,,of the complement event.
Dialogue: 0,0:18:40.82,0:18:42.25,Default,,0000,0000,0000,,So this is important.
Dialogue: 0,0:18:42.25,0:18:45.98,Default,,0000,0000,0000,,It's a very handy tool\Nto have in your tool box.
Dialogue: 0,0:18:45.98,0:18:47.35,Default,,0000,0000,0000,,For example, if\NI ask you to find
Dialogue: 0,0:18:47.35,0:18:50.17,Default,,0000,0000,0000,,the probability of not\Nrolling a sum of 11
Dialogue: 0,0:18:50.17,0:18:52.39,Default,,0000,0000,0000,,when you're rolling two dice--
Dialogue: 0,0:18:52.39,0:18:54.33,Default,,0000,0000,0000,,again, you'd build your table--
Dialogue: 0,0:18:54.33,0:18:56.80,Default,,0000,0000,0000,,the probability of\Nnot rolling an 11
Dialogue: 0,0:18:56.80,0:19:02.71,Default,,0000,0000,0000,,is the same thing as 1 minus the\Nprobability of rolling an 11.
Dialogue: 0,0:19:02.71,0:19:06.22,Default,,0000,0000,0000,,The probability of the\Ncomplement of something
Dialogue: 0,0:19:06.22,0:19:08.87,Default,,0000,0000,0000,,is 1 minus the probability\Nof that something.
Dialogue: 0,0:19:08.87,0:19:12.23,Default,,0000,0000,0000,,So it's very easy to calculate\Nthe probability if sum is 11,
Dialogue: 0,0:19:12.23,0:19:13.94,Default,,0000,0000,0000,,because there are only\Ntwo possibilities.
Dialogue: 0,0:19:13.94,0:19:15.72,Default,,0000,0000,0000,,You can get 6, 5 or 5, 6.
Dialogue: 0,0:19:15.72,0:19:17.95,Default,,0000,0000,0000,,So the probability of\Ngetting a sum of 11
Dialogue: 0,0:19:17.95,0:19:20.66,Default,,0000,0000,0000,,is simply 2 out of 36.
Dialogue: 0,0:19:20.66,0:19:24.16,Default,,0000,0000,0000,,So the probability of\Nnot getting a sum of 11
Dialogue: 0,0:19:24.16,0:19:28.15,Default,,0000,0000,0000,,is 1 minus 2 out of 36,\Nwhich would be 34 out of 306.
Dialogue: 0,0:19:28.15,0:19:31.54,Default,,0000,0000,0000,,Complement rule is a very\Npowerful tool in your tool box.
Dialogue: 0,0:19:31.54,0:19:35.58,Default,,0000,0000,0000,,If you wanted to reduce that\Nto 17/18, you're free to.
Dialogue: 0,0:19:35.58,0:19:38.86,Default,,0000,0000,0000,,I want to talk a little bit\Nabout the English language
Dialogue: 0,0:19:38.86,0:19:40.52,Default,,0000,0000,0000,,as relates to this.
Dialogue: 0,0:19:40.52,0:19:42.77,Default,,0000,0000,0000,,It turns out that when\Nwe're doing probabilities,
Dialogue: 0,0:19:42.77,0:19:44.86,Default,,0000,0000,0000,,we tend to come\Nto these problems
Dialogue: 0,0:19:44.86,0:19:47.50,Default,,0000,0000,0000,,where we're asked to find\Nthe probability of at least
Dialogue: 0,0:19:47.50,0:19:50.56,Default,,0000,0000,0000,,one something-- and I want to\Ntalk a little bit about that
Dialogue: 0,0:19:50.56,0:19:54.47,Default,,0000,0000,0000,,in particular, so that when you\Nsee it, it won't be confusing.
Dialogue: 0,0:19:54.47,0:19:58.01,Default,,0000,0000,0000,,For example, if I roll a\Nstandard die three times,
Dialogue: 0,0:19:58.01,0:20:00.64,Default,,0000,0000,0000,,what's the probability\Nwe get at least one 4?
Dialogue: 0,0:20:00.64,0:20:02.98,Default,,0000,0000,0000,,You see that at least\None quite often.
Dialogue: 0,0:20:02.98,0:20:04.54,Default,,0000,0000,0000,,At least one 4.
Dialogue: 0,0:20:04.54,0:20:05.74,Default,,0000,0000,0000,,We'll think about it.
Dialogue: 0,0:20:05.74,0:20:09.34,Default,,0000,0000,0000,,At least one 4 means\Nyou get exactly one 4,
Dialogue: 0,0:20:09.34,0:20:12.73,Default,,0000,0000,0000,,or exactly two 4's\Nor exactly three 4's.
Dialogue: 0,0:20:12.73,0:20:17.23,Default,,0000,0000,0000,,That's what it means to\Nhave rolled at least one 4
Dialogue: 0,0:20:17.23,0:20:21.18,Default,,0000,0000,0000,,in the context of rolling\Nthe die three times.
Dialogue: 0,0:20:21.18,0:20:22.98,Default,,0000,0000,0000,,One way to answer\Nthis question is just
Dialogue: 0,0:20:22.98,0:20:25.20,Default,,0000,0000,0000,,to do three separate\Nprobability calculations
Dialogue: 0,0:20:25.20,0:20:26.37,Default,,0000,0000,0000,,and just add at the results.
Dialogue: 0,0:20:26.37,0:20:29.19,Default,,0000,0000,0000,,Just find out the probability\Nof getting exactly one 4.
Dialogue: 0,0:20:29.19,0:20:32.79,Default,,0000,0000,0000,,Then find the probability\Nof getting exactly two 4's.
Dialogue: 0,0:20:32.79,0:20:35.49,Default,,0000,0000,0000,,Then find the probability of\Ngetting exactly three 4's.
Dialogue: 0,0:20:35.49,0:20:37.99,Default,,0000,0000,0000,,And just add them all up.
Dialogue: 0,0:20:37.99,0:20:39.85,Default,,0000,0000,0000,,That's one way to do it.
Dialogue: 0,0:20:39.85,0:20:44.27,Default,,0000,0000,0000,,You could take this\Nand this and this,
Dialogue: 0,0:20:44.27,0:20:48.29,Default,,0000,0000,0000,,and just add up the three\Nseparate probabilities.
Dialogue: 0,0:20:48.29,0:20:50.50,Default,,0000,0000,0000,,That's quite a bit of work.
Dialogue: 0,0:20:50.50,0:20:52.86,Default,,0000,0000,0000,,There is a better way.
Dialogue: 0,0:20:52.86,0:20:56.52,Default,,0000,0000,0000,,A better way is to use\Nthe complement rule,
Dialogue: 0,0:20:56.52,0:20:58.99,Default,,0000,0000,0000,,because if we roll\Nthe die three times,
Dialogue: 0,0:20:58.99,0:21:00.87,Default,,0000,0000,0000,,there are only four\Npossibilities altogether.
Dialogue: 0,0:21:00.87,0:21:03.13,Default,,0000,0000,0000,,We've talked about\Nthree of them.
Dialogue: 0,0:21:03.13,0:21:08.10,Default,,0000,0000,0000,,We've talked about getting\None 4 or two 4's or three 4's.
Dialogue: 0,0:21:08.10,0:21:09.45,Default,,0000,0000,0000,,There's only one other way.
Dialogue: 0,0:21:09.45,0:21:12.29,Default,,0000,0000,0000,,And that's to get no 4's at all.
Dialogue: 0,0:21:12.29,0:21:14.04,Default,,0000,0000,0000,,So wouldn't it be much\Neasier to calculate
Dialogue: 0,0:21:14.04,0:21:17.01,Default,,0000,0000,0000,,the probability of\Nnot getting any fours
Dialogue: 0,0:21:17.01,0:21:19.83,Default,,0000,0000,0000,,and then say 1 minus that?
Dialogue: 0,0:21:19.83,0:21:22.69,Default,,0000,0000,0000,,And that's what I'm going to do.
Dialogue: 0,0:21:22.69,0:21:24.36,Default,,0000,0000,0000,,The complement rule\Nsays the probability
Dialogue: 0,0:21:24.36,0:21:27.68,Default,,0000,0000,0000,,of getting at least one 4\Nis 1 minus the probability
Dialogue: 0,0:21:27.68,0:21:30.40,Default,,0000,0000,0000,,of getting no 4's,\Nbecause getting no 4's is
Dialogue: 0,0:21:30.40,0:21:33.62,Default,,0000,0000,0000,,the complementary event\Nof getting at least one 4.
Dialogue: 0,0:21:33.62,0:21:38.65,Default,,0000,0000,0000,,And getting no 4's is a much\Neasier probability calculation.
Dialogue: 0,0:21:38.65,0:21:42.45,Default,,0000,0000,0000,,So how did we calculate the\Nprobability of getting no 4's?
Dialogue: 0,0:21:42.45,0:21:45.50,Default,,0000,0000,0000,,And this comes back to the\Nstuff we talked about earlier.
Dialogue: 0,0:21:45.50,0:21:47.13,Default,,0000,0000,0000,,Remember when we\Nstarted studying this,
Dialogue: 0,0:21:47.13,0:21:50.13,Default,,0000,0000,0000,,I asked you if you\Nknow how to count.
Dialogue: 0,0:21:50.13,0:21:52.02,Default,,0000,0000,0000,,The solution that\Nproblem requires
Dialogue: 0,0:21:52.02,0:21:55.56,Default,,0000,0000,0000,,us to count using some of\Nour combinatorial formulas.
Dialogue: 0,0:21:55.56,0:21:58.84,Default,,0000,0000,0000,,And that's why we studied\Nthem, so we could use them.
Dialogue: 0,0:21:58.84,0:22:00.87,Default,,0000,0000,0000,,So if a standard die\Nis rolled three times,
Dialogue: 0,0:22:00.87,0:22:03.87,Default,,0000,0000,0000,,what is the probability\Nthat we get at least one 4?
Dialogue: 0,0:22:03.87,0:22:06.75,Default,,0000,0000,0000,,Well, we just talked about\Nthe easiest way to do it is
Dialogue: 0,0:22:06.75,0:22:09.45,Default,,0000,0000,0000,,to calculate the probability\Nof getting at least one 4,
Dialogue: 0,0:22:09.45,0:22:12.48,Default,,0000,0000,0000,,as 1 minus the probability\Nthat we don't get any 4's.
Dialogue: 0,0:22:12.48,0:22:16.08,Default,,0000,0000,0000,,That probability of getting\Nno 4's is the key calculation
Dialogue: 0,0:22:16.08,0:22:17.79,Default,,0000,0000,0000,,in the whole problem,\Nonce you understand
Dialogue: 0,0:22:17.79,0:22:18.83,Default,,0000,0000,0000,,what you're trying to do.
Dialogue: 0,0:22:18.83,0:22:21.27,Default,,0000,0000,0000,,So how do you calculate the\Nprobability you don't get any
Dialogue: 0,0:22:21.27,0:22:22.07,Default,,0000,0000,0000,,4's?
Dialogue: 0,0:22:22.07,0:22:23.53,Default,,0000,0000,0000,,Well, I would think\Nof it this way.
Dialogue: 0,0:22:23.53,0:22:26.01,Default,,0000,0000,0000,,The probability of getting\Nno 4's is the number
Dialogue: 0,0:22:26.01,0:22:29.40,Default,,0000,0000,0000,,of ways to get anything but a\N4 divided by the number of ways
Dialogue: 0,0:22:29.40,0:22:31.47,Default,,0000,0000,0000,,to get anything with\Nyour three rolls.
Dialogue: 0,0:22:31.47,0:22:34.47,Default,,0000,0000,0000,,I think I'll start with\Nthe denominator first.
Dialogue: 0,0:22:34.47,0:22:37.26,Default,,0000,0000,0000,,Let's find the number of\Nways to get any three rolls.
Dialogue: 0,0:22:37.26,0:22:39.66,Default,,0000,0000,0000,,Going all the way back to\Nthe counting principle,
Dialogue: 0,0:22:39.66,0:22:42.67,Default,,0000,0000,0000,,if you're looking for the number\Nof ways to get any three rolls,
Dialogue: 0,0:22:42.67,0:22:46.26,Default,,0000,0000,0000,,remember each roll can\Nbe a 1, 2, 3, 4, 5, or 6.
Dialogue: 0,0:22:46.26,0:22:48.14,Default,,0000,0000,0000,,Think of this as a\Nthree-stage experiment.
Dialogue: 0,0:22:48.14,0:22:50.64,Default,,0000,0000,0000,,You can look at how many ways\Nyou can get something for roll
Dialogue: 0,0:22:50.64,0:22:53.04,Default,,0000,0000,0000,,one, row two, roll three.
Dialogue: 0,0:22:53.04,0:22:55.69,Default,,0000,0000,0000,,The counting principle says\Nto multiply them together.
Dialogue: 0,0:22:55.69,0:22:58.08,Default,,0000,0000,0000,,So the number of possibilities\Nfor the first roll-
Dialogue: 0,0:22:58.08,0:22:59.94,Default,,0000,0000,0000,,well, there are\Nsix possibilities.
Dialogue: 0,0:22:59.94,0:23:02.55,Default,,0000,0000,0000,,As for the second, there\Nare still six possibilities.
Dialogue: 0,0:23:02.55,0:23:04.95,Default,,0000,0000,0000,,And for the third, there\Nare still six possibilities.
Dialogue: 0,0:23:04.95,0:23:10.20,Default,,0000,0000,0000,,So if the counting principle\Nsays the possible die rolls
Dialogue: 0,0:23:10.20,0:23:13.83,Default,,0000,0000,0000,,you can get when you roll three\Ntimes is simply 6 times 6 times
Dialogue: 0,0:23:13.83,0:23:16.78,Default,,0000,0000,0000,,6, which happens to be 216.
Dialogue: 0,0:23:16.78,0:23:18.83,Default,,0000,0000,0000,,That's the denominator.
Dialogue: 0,0:23:18.83,0:23:21.97,Default,,0000,0000,0000,,But what about the number of\Nways to get anything but a 4?
Dialogue: 0,0:23:21.97,0:23:26.22,Default,,0000,0000,0000,,Well, after having done the\Ncalculation we just did,
Dialogue: 0,0:23:26.22,0:23:27.68,Default,,0000,0000,0000,,I think you'll see\Nthat it's pretty
Dialogue: 0,0:23:27.68,0:23:30.77,Default,,0000,0000,0000,,easy to look at the\Nnumerator and see that it's
Dialogue: 0,0:23:30.77,0:23:33.47,Default,,0000,0000,0000,,the same thought process.
Dialogue: 0,0:23:33.47,0:23:35.78,Default,,0000,0000,0000,,But this time we're not\Nallowing the possibility
Dialogue: 0,0:23:35.78,0:23:38.48,Default,,0000,0000,0000,,of getting a 4, because we're\Nlooking for the number of ways
Dialogue: 0,0:23:38.48,0:23:40.34,Default,,0000,0000,0000,,to get anything but a 4.
Dialogue: 0,0:23:40.34,0:23:42.95,Default,,0000,0000,0000,,So we knocked the 4 out\Nas being a possibility,
Dialogue: 0,0:23:42.95,0:23:44.70,Default,,0000,0000,0000,,but the process stays the same.
Dialogue: 0,0:23:44.70,0:23:46.13,Default,,0000,0000,0000,,We're still doing\Nthree die rolls.
Dialogue: 0,0:23:46.13,0:23:47.63,Default,,0000,0000,0000,,We just don't want any 4's.
Dialogue: 0,0:23:47.63,0:23:49.76,Default,,0000,0000,0000,,So there's only\Nfive possibilities
Dialogue: 0,0:23:49.76,0:23:52.70,Default,,0000,0000,0000,,for the first die roll if\Nyou don't want it to be a 4.
Dialogue: 0,0:23:52.70,0:23:54.92,Default,,0000,0000,0000,,There are only five\Npossibilities for the die roll
Dialogue: 0,0:23:54.92,0:23:57.16,Default,,0000,0000,0000,,two if you don't\Nwant it to be a 4.
Dialogue: 0,0:23:57.16,0:23:58.70,Default,,0000,0000,0000,,And there are only\Nfive possibilities
Dialogue: 0,0:23:58.70,0:24:01.67,Default,,0000,0000,0000,,for die roll three if you\Ndon't want it to be a 4.
Dialogue: 0,0:24:01.67,0:24:04.86,Default,,0000,0000,0000,,5 times 5 times 5 is 125.
Dialogue: 0,0:24:04.86,0:24:11.72,Default,,0000,0000,0000,,So the probability of getting\Nno 4's is simply 125 over 216.
Dialogue: 0,0:24:11.72,0:24:13.19,Default,,0000,0000,0000,,But that wasn't\Nthe final answer.
Dialogue: 0,0:24:13.19,0:24:16.07,Default,,0000,0000,0000,,We want to know the probability\Nof getting at least one 4.
Dialogue: 0,0:24:16.07,0:24:18.62,Default,,0000,0000,0000,,The key calculation was getting\Nthe probability of no fours,
Dialogue: 0,0:24:18.62,0:24:20.36,Default,,0000,0000,0000,,but it wasn't the final answer.
Dialogue: 0,0:24:20.36,0:24:28.68,Default,,0000,0000,0000,,The probability is 1 minus that,\Nwhich would be 91 over 216.
Dialogue: 0,0:24:28.68,0:24:29.31,Default,,0000,0000,0000,,How about this?
Dialogue: 0,0:24:29.31,0:24:31.17,Default,,0000,0000,0000,,If you draw three cards\Nfrom a standard deck
Dialogue: 0,0:24:31.17,0:24:34.83,Default,,0000,0000,0000,,without replacement, what is the\Nprobability that at least one
Dialogue: 0,0:24:34.83,0:24:36.09,Default,,0000,0000,0000,,is a face card?
Dialogue: 0,0:24:36.09,0:24:39.42,Default,,0000,0000,0000,,Write your answers as a percent\Nrounded to one decimal place.
Dialogue: 0,0:24:39.42,0:24:40.80,Default,,0000,0000,0000,,So as we learned,\Nthe probability
Dialogue: 0,0:24:40.80,0:24:43.53,Default,,0000,0000,0000,,of at least one something\Nis 1 minus the probability
Dialogue: 0,0:24:43.53,0:24:45.10,Default,,0000,0000,0000,,of none of that thing.
Dialogue: 0,0:24:45.10,0:24:46.47,Default,,0000,0000,0000,,So in particular,\Nthe probability
Dialogue: 0,0:24:46.47,0:24:49.56,Default,,0000,0000,0000,,of at least one face card\Nis 1 minus the probability
Dialogue: 0,0:24:49.56,0:24:52.35,Default,,0000,0000,0000,,of no face cards,\Nor zero face cards.
Dialogue: 0,0:24:52.35,0:24:55.95,Default,,0000,0000,0000,,And as we've seen before, that\Nprobability of zero face cards
Dialogue: 0,0:24:55.95,0:24:59.83,Default,,0000,0000,0000,,is the key calculation.
Dialogue: 0,0:24:59.83,0:25:02.94,Default,,0000,0000,0000,,Well, what is the\Nprobability of no face cards?
Dialogue: 0,0:25:02.94,0:25:04.83,Default,,0000,0000,0000,,There are 12 face\Ncards in the deck.
Dialogue: 0,0:25:04.83,0:25:06.74,Default,,0000,0000,0000,,We've done this multiple times.
Dialogue: 0,0:25:06.74,0:25:10.10,Default,,0000,0000,0000,,So that means there are 40\Ncards that aren't face cards,
Dialogue: 0,0:25:10.10,0:25:12.53,Default,,0000,0000,0000,,because 52 minus 12 is 40.
Dialogue: 0,0:25:12.53,0:25:14.81,Default,,0000,0000,0000,,So we think of partitioning\Nthe deck into face
Dialogue: 0,0:25:14.81,0:25:16.70,Default,,0000,0000,0000,,cards and non-face cards.
Dialogue: 0,0:25:16.70,0:25:17.93,Default,,0000,0000,0000,,There are 12 face cards.
Dialogue: 0,0:25:17.93,0:25:20.69,Default,,0000,0000,0000,,There 40 non-face cards.
Dialogue: 0,0:25:20.69,0:25:23.36,Default,,0000,0000,0000,,So the probability you\Ndon't get a face card
Dialogue: 0,0:25:23.36,0:25:25.60,Default,,0000,0000,0000,,is the number of ways\Nnot to get a face card--
Dialogue: 0,0:25:25.60,0:25:27.02,Default,,0000,0000,0000,,in other words,\Nthe number of ways
Dialogue: 0,0:25:27.02,0:25:29.54,Default,,0000,0000,0000,,to get anything\Nbut a face card--
Dialogue: 0,0:25:29.54,0:25:31.82,Default,,0000,0000,0000,,divided by the number of\Nways to get any three cards.
Dialogue: 0,0:25:31.82,0:25:35.36,Default,,0000,0000,0000,,
Dialogue: 0,0:25:35.36,0:25:38.82,Default,,0000,0000,0000,,You've done these problems\Nin earlier sections.
Dialogue: 0,0:25:38.82,0:25:42.50,Default,,0000,0000,0000,,When we're drawing face cards,\Nwe don't care about order.
Dialogue: 0,0:25:42.50,0:25:44.37,Default,,0000,0000,0000,,Because when you pull\Na card out of the deck,
Dialogue: 0,0:25:44.37,0:25:47.09,Default,,0000,0000,0000,,you put it in your hand, it\Ndoesn't matter what order
Dialogue: 0,0:25:47.09,0:25:48.96,Default,,0000,0000,0000,,the card came into your hand.
Dialogue: 0,0:25:48.96,0:25:51.80,Default,,0000,0000,0000,,So this is a combination.
Dialogue: 0,0:25:51.80,0:25:53.54,Default,,0000,0000,0000,,If we're trying not\Nto get a face card,
Dialogue: 0,0:25:53.54,0:25:56.56,Default,,0000,0000,0000,,we've got to draw from the 40\Nthings that aren't face cards.
Dialogue: 0,0:25:56.56,0:25:58.88,Default,,0000,0000,0000,,So it's a combination\Nof 40 things,
Dialogue: 0,0:25:58.88,0:26:00.38,Default,,0000,0000,0000,,and we're choosing three cards.
Dialogue: 0,0:26:00.38,0:26:03.92,Default,,0000,0000,0000,,So it's a combination of 40\Nthings, taken three at a time.
Dialogue: 0,0:26:03.92,0:26:08.06,Default,,0000,0000,0000,,Down bottom, when we don't\Ncare what the cards are,
Dialogue: 0,0:26:08.06,0:26:09.59,Default,,0000,0000,0000,,it can be any 52.
Dialogue: 0,0:26:09.59,0:26:11.42,Default,,0000,0000,0000,,But we're still\Ndrawing three cards.
Dialogue: 0,0:26:11.42,0:26:14.74,Default,,0000,0000,0000,,
Dialogue: 0,0:26:14.74,0:26:16.32,Default,,0000,0000,0000,,So it's a combination\Nof 40 things,
Dialogue: 0,0:26:16.32,0:26:18.70,Default,,0000,0000,0000,,taking three at a time-- that's\Nthe number of ways to get
Dialogue: 0,0:26:18.70,0:26:20.32,Default,,0000,0000,0000,,anything but a face card--
Dialogue: 0,0:26:20.32,0:26:23.83,Default,,0000,0000,0000,,over a combination of 52\Nthings taken three at a time.
Dialogue: 0,0:26:23.83,0:26:25.75,Default,,0000,0000,0000,,That's when you don't\Ncare what the cards are.
Dialogue: 0,0:26:25.75,0:26:26.62,Default,,0000,0000,0000,,You're drawing three.
Dialogue: 0,0:26:26.62,0:26:28.65,Default,,0000,0000,0000,,But you don't care\Nwhat they are.
Dialogue: 0,0:26:28.65,0:26:30.65,Default,,0000,0000,0000,,And in case you really\Nhave forgotten your shift
Dialogue: 0,0:26:30.65,0:26:36.01,Default,,0000,0000,0000,,combinations to get\Nyour calculator answer,
Dialogue: 0,0:26:36.01,0:26:37.86,Default,,0000,0000,0000,,I put them there for you.
Dialogue: 0,0:26:37.86,0:26:39.69,Default,,0000,0000,0000,,I hope you didn't need it.
Dialogue: 0,0:26:39.69,0:26:43.03,Default,,0000,0000,0000,,They ask you to do a\Ndecimal approximation.
Dialogue: 0,0:26:43.03,0:26:45.15,Default,,0000,0000,0000,,So you need to change\Nthat now to a decimal.
Dialogue: 0,0:26:45.15,0:26:49.29,Default,,0000,0000,0000,,That's 0.44706.
Dialogue: 0,0:26:49.29,0:26:51.40,Default,,0000,0000,0000,,They ask you to change\Nit to a percent rounded
Dialogue: 0,0:26:51.40,0:26:52.56,Default,,0000,0000,0000,,to one decimal place.
Dialogue: 0,0:26:52.56,0:26:55.71,Default,,0000,0000,0000,,So take plenty of decimals,\Nbecause you have to change it
Dialogue: 0,0:26:55.71,0:26:58.58,Default,,0000,0000,0000,,to a percent and then round it.
Dialogue: 0,0:26:58.58,0:27:01.64,Default,,0000,0000,0000,,So do not just write 0.44.
Dialogue: 0,0:27:01.64,0:27:03.17,Default,,0000,0000,0000,,Write several decimal places.
Dialogue: 0,0:27:03.17,0:27:07.48,Default,,0000,0000,0000,,So it's 0.44706 to\Nseveral decimal places.
Dialogue: 0,0:27:07.48,0:27:09.21,Default,,0000,0000,0000,,So you continue\Nyour calculation.
Dialogue: 0,0:27:09.21,0:27:12.27,Default,,0000,0000,0000,,The probability of getting\Nat least one face card
Dialogue: 0,0:27:12.27,0:27:15.18,Default,,0000,0000,0000,,is 1 minus the probability\Nof no face cards, which
Dialogue: 0,0:27:15.18,0:27:18.30,Default,,0000,0000,0000,,we've calculated to be 0.44706.
Dialogue: 0,0:27:18.30,0:27:21.44,Default,,0000,0000,0000,,Notice I'm\Nemphasizing yet again,
Dialogue: 0,0:27:21.44,0:27:24.45,Default,,0000,0000,0000,,carry plenty of decimal places\Nduring this calculation.
Dialogue: 0,0:27:24.45,0:27:27.90,Default,,0000,0000,0000,,Round off in the very last step.
Dialogue: 0,0:27:27.90,0:27:28.92,Default,,0000,0000,0000,,So do that subtraction.
Dialogue: 0,0:27:28.92,0:27:32.64,Default,,0000,0000,0000,,You get 0.55294.
Dialogue: 0,0:27:32.64,0:27:34.50,Default,,0000,0000,0000,,You're still not ready\Nto round, because it
Dialogue: 0,0:27:34.50,0:27:38.07,Default,,0000,0000,0000,,says to give your answer\Nas a percent rounded
Dialogue: 0,0:27:38.07,0:27:39.03,Default,,0000,0000,0000,,to one decimal place.
Dialogue: 0,0:27:39.03,0:27:41.76,Default,,0000,0000,0000,,So you first change it to\Npercent, and then you round.
Dialogue: 0,0:27:41.76,0:27:43.89,Default,,0000,0000,0000,,You sort of see now\Nwhy I'm emphasizing
Dialogue: 0,0:27:43.89,0:27:46.99,Default,,0000,0000,0000,,how important it is to carry\Nplenty of decimal places.
Dialogue: 0,0:27:46.99,0:27:48.63,Default,,0000,0000,0000,,Now change it to\Npercent, which means
Dialogue: 0,0:27:48.63,0:27:51.66,Default,,0000,0000,0000,,you move the decimal place\Ntwo places to the right.
Dialogue: 0,0:27:51.66,0:27:55.74,Default,,0000,0000,0000,,And finally, you round\Nthat to one decimal place.
Dialogue: 0,0:27:55.74,0:27:58.20,Default,,0000,0000,0000,,Suppose we take the\Nproblem that we just solved
Dialogue: 0,0:27:58.20,0:28:01.05,Default,,0000,0000,0000,,and change it ever so\Nslightly so that now we're
Dialogue: 0,0:28:01.05,0:28:03.14,Default,,0000,0000,0000,,drawing three cards\Nfrom a standard deck,
Dialogue: 0,0:28:03.14,0:28:05.23,Default,,0000,0000,0000,,but we're drawing\Nwith replacement.
Dialogue: 0,0:28:05.23,0:28:08.10,Default,,0000,0000,0000,,We still want to know what is\Nthe probability that at least
Dialogue: 0,0:28:08.10,0:28:09.66,Default,,0000,0000,0000,,one is a face card.
Dialogue: 0,0:28:09.66,0:28:11.46,Default,,0000,0000,0000,,And we'll take our\Nanswer as a percent
Dialogue: 0,0:28:11.46,0:28:12.69,Default,,0000,0000,0000,,rounded to one decimal place.
Dialogue: 0,0:28:12.69,0:28:14.58,Default,,0000,0000,0000,,So the only difference\Nhere is that we're
Dialogue: 0,0:28:14.58,0:28:15.96,Default,,0000,0000,0000,,drawing with replacement.
Dialogue: 0,0:28:15.96,0:28:19.36,Default,,0000,0000,0000,,How does that change everything?
Dialogue: 0,0:28:19.36,0:28:22.01,Default,,0000,0000,0000,,What's different about it?
Dialogue: 0,0:28:22.01,0:28:23.75,Default,,0000,0000,0000,,Because we're drawing\Nwith replacement,
Dialogue: 0,0:28:23.75,0:28:26.75,Default,,0000,0000,0000,,this does not involve\Ncounting things
Dialogue: 0,0:28:26.75,0:28:29.22,Default,,0000,0000,0000,,using either the permutation\Nor combination formulas.
Dialogue: 0,0:28:29.22,0:28:31.67,Default,,0000,0000,0000,,Those are what we use when\Nwe have distinct objects,
Dialogue: 0,0:28:31.67,0:28:33.91,Default,,0000,0000,0000,,and we're looking at\Nhow many arrangements--
Dialogue: 0,0:28:33.91,0:28:35.66,Default,,0000,0000,0000,,or how many ways we\Ncan choose from those.
Dialogue: 0,0:28:35.66,0:28:38.30,Default,,0000,0000,0000,,When we're drawing\Nwith replacement,
Dialogue: 0,0:28:38.30,0:28:39.71,Default,,0000,0000,0000,,it just doesn't fit that.
Dialogue: 0,0:28:39.71,0:28:42.74,Default,,0000,0000,0000,,So we have to go back to the\Ncounting principle for problems
Dialogue: 0,0:28:42.74,0:28:43.55,Default,,0000,0000,0000,,like this.
Dialogue: 0,0:28:43.55,0:28:45.77,Default,,0000,0000,0000,,And that's the difference\Nbetween this problem
Dialogue: 0,0:28:45.77,0:28:48.40,Default,,0000,0000,0000,,and the one you just solved.
Dialogue: 0,0:28:48.40,0:28:52.27,Default,,0000,0000,0000,,But still, with the\Ncomplement rule applies--
Dialogue: 0,0:28:52.27,0:28:54.07,Default,,0000,0000,0000,,and we can calculate\Nthe probability
Dialogue: 0,0:28:54.07,0:28:58.30,Default,,0000,0000,0000,,there's at least one face\Ncard by calculating 1
Dialogue: 0,0:28:58.30,0:29:01.09,Default,,0000,0000,0000,,minus the probability that\Nthere aren't any face cards,
Dialogue: 0,0:29:01.09,0:29:02.84,Default,,0000,0000,0000,,or that there is\Nzero face cards.
Dialogue: 0,0:29:02.84,0:29:06.63,Default,,0000,0000,0000,,So this part doesn't\Nchange at all.
Dialogue: 0,0:29:06.63,0:29:09.42,Default,,0000,0000,0000,,And that calculation of\Ngetting zero face cards
Dialogue: 0,0:29:09.42,0:29:10.99,Default,,0000,0000,0000,,is really the key\Nto the whole thing.
Dialogue: 0,0:29:10.99,0:29:13.80,Default,,0000,0000,0000,,Because once we get that, the\Nfinal answer will be just 1
Dialogue: 0,0:29:13.80,0:29:15.03,Default,,0000,0000,0000,,minus that answer.
Dialogue: 0,0:29:15.03,0:29:18.91,Default,,0000,0000,0000,,
Dialogue: 0,0:29:18.91,0:29:21.33,Default,,0000,0000,0000,,Remember also, there\Nare 12 face cards.
Dialogue: 0,0:29:21.33,0:29:25.20,Default,,0000,0000,0000,,So if we're trying to\Nnot draw a face card,
Dialogue: 0,0:29:25.20,0:29:27.42,Default,,0000,0000,0000,,we're looking for one\Nof the non-face cards.
Dialogue: 0,0:29:27.42,0:29:32.39,Default,,0000,0000,0000,,And there are 40 of them,\Nbecause 52 minus 12 is 40.
Dialogue: 0,0:29:32.39,0:29:35.99,Default,,0000,0000,0000,,And as before, the number of\Nways you can get no face cards
Dialogue: 0,0:29:35.99,0:29:39.74,Default,,0000,0000,0000,,is simply the number of ways\Nthat you don't get a face card
Dialogue: 0,0:29:39.74,0:29:43.64,Default,,0000,0000,0000,,when you're drawing those three\Ncard over any possible way you
Dialogue: 0,0:29:43.64,0:29:46.04,Default,,0000,0000,0000,,could draw three\Ncards from the deck.
Dialogue: 0,0:29:46.04,0:29:50.32,Default,,0000,0000,0000,,And again, this time\Nwith replacement.
Dialogue: 0,0:29:50.32,0:29:54.73,Default,,0000,0000,0000,,So the answer for the zero\Nface cards probability
Dialogue: 0,0:29:54.73,0:29:55.63,Default,,0000,0000,0000,,will be a fraction.
Dialogue: 0,0:29:55.63,0:29:58.22,Default,,0000,0000,0000,,And we have to calculate the\Nnumerator and denominator,
Dialogue: 0,0:29:58.22,0:30:01.22,Default,,0000,0000,0000,,and we're home free.
Dialogue: 0,0:30:01.22,0:30:04.19,Default,,0000,0000,0000,,But as I said, when we're\Ndoing it with replacement,
Dialogue: 0,0:30:04.19,0:30:07.19,Default,,0000,0000,0000,,we can't calculate it with\Nour combination formulas.
Dialogue: 0,0:30:07.19,0:30:10.46,Default,,0000,0000,0000,,We have to go back to\Nthe counting principle.
Dialogue: 0,0:30:10.46,0:30:12.11,Default,,0000,0000,0000,,And here it says\Nwe're just doing
Dialogue: 0,0:30:12.11,0:30:14.84,Default,,0000,0000,0000,,three events, three\Nsub-tasks-- however
Dialogue: 0,0:30:14.84,0:30:16.32,Default,,0000,0000,0000,,you want to think of it.
Dialogue: 0,0:30:16.32,0:30:17.69,Default,,0000,0000,0000,,And we've got to\Ncount the number
Dialogue: 0,0:30:17.69,0:30:19.35,Default,,0000,0000,0000,,of ways you can do each one.
Dialogue: 0,0:30:19.35,0:30:22.70,Default,,0000,0000,0000,,Well, if you're trying\Nto not draw face card,
Dialogue: 0,0:30:22.70,0:30:25.22,Default,,0000,0000,0000,,on the first draw,\Nthere are 40 ways
Dialogue: 0,0:30:25.22,0:30:27.11,Default,,0000,0000,0000,,of not getting a face card.
Dialogue: 0,0:30:27.11,0:30:31.24,Default,,0000,0000,0000,,And since you're putting\Nthe card back in the deck,
Dialogue: 0,0:30:31.24,0:30:35.35,Default,,0000,0000,0000,,there are 40 ways to not choose\Na face card on the second draw.
Dialogue: 0,0:30:35.35,0:30:38.38,Default,,0000,0000,0000,,And again, because you're\Nputting the card back,
Dialogue: 0,0:30:38.38,0:30:42.13,Default,,0000,0000,0000,,there are 40 ways not to choose\Na face card the third time.
Dialogue: 0,0:30:42.13,0:30:45.28,Default,,0000,0000,0000,,So the numerator will\Nbe 40 times 40 times 40.
Dialogue: 0,0:30:45.28,0:30:46.58,Default,,0000,0000,0000,,How about the denominator?
Dialogue: 0,0:30:46.58,0:30:49.54,Default,,0000,0000,0000,,Well, same thought process,\Nbut in the denominator,
Dialogue: 0,0:30:49.54,0:30:51.91,Default,,0000,0000,0000,,you don't care what\Nit comes out to be.
Dialogue: 0,0:30:51.91,0:30:53.87,Default,,0000,0000,0000,,It could be any of the 52 cards.
Dialogue: 0,0:30:53.87,0:30:55.87,Default,,0000,0000,0000,,And again, because you're\Nputting it back in,
Dialogue: 0,0:30:55.87,0:30:58.09,Default,,0000,0000,0000,,you still have 52\Ncards to choose from.
Dialogue: 0,0:30:58.09,0:30:59.95,Default,,0000,0000,0000,,You don't care what the card is.
Dialogue: 0,0:30:59.95,0:31:01.72,Default,,0000,0000,0000,,And the third draw is the same.
Dialogue: 0,0:31:01.72,0:31:04.18,Default,,0000,0000,0000,,You have all 52\Ncards to choose from.
Dialogue: 0,0:31:04.18,0:31:07.42,Default,,0000,0000,0000,,And you don't care\Nwhat the card is.
Dialogue: 0,0:31:07.42,0:31:11.05,Default,,0000,0000,0000,,So the result will be 40 times\N40 times 40 over 52 times
Dialogue: 0,0:31:11.05,0:31:12.76,Default,,0000,0000,0000,,52 times 52.
Dialogue: 0,0:31:12.76,0:31:18.37,Default,,0000,0000,0000,,That's 64,000 over 140,608.
Dialogue: 0,0:31:18.37,0:31:23.47,Default,,0000,0000,0000,,Now, the answer they want is\Ndecimals rounded to one place.
Dialogue: 0,0:31:23.47,0:31:24.52,Default,,0000,0000,0000,,And that's a percent.
Dialogue: 0,0:31:24.52,0:31:26.08,Default,,0000,0000,0000,,So we need to\Nchange that fraction
Dialogue: 0,0:31:26.08,0:31:27.64,Default,,0000,0000,0000,,to a decimal at some point.
Dialogue: 0,0:31:27.64,0:31:29.26,Default,,0000,0000,0000,,You might as well do it now.
Dialogue: 0,0:31:29.26,0:31:31.54,Default,,0000,0000,0000,,Always take quite a\Nfew more decimal places
Dialogue: 0,0:31:31.54,0:31:33.04,Default,,0000,0000,0000,,than you think\Nyou're going to need.
Dialogue: 0,0:31:33.04,0:31:34.79,Default,,0000,0000,0000,,It says the round it\Nto one decimal place.
Dialogue: 0,0:31:34.79,0:31:38.14,Default,,0000,0000,0000,,But it also says to\Nwrite as a percent.
Dialogue: 0,0:31:38.14,0:31:40.90,Default,,0000,0000,0000,,Writing it as a percent causes\Nyou to move the decimal place
Dialogue: 0,0:31:40.90,0:31:42.68,Default,,0000,0000,0000,,two places to the right.
Dialogue: 0,0:31:42.68,0:31:44.68,Default,,0000,0000,0000,,Then you want another\Ndecimal place of accuracy.
Dialogue: 0,0:31:44.68,0:31:45.70,Default,,0000,0000,0000,,That's three.
Dialogue: 0,0:31:45.70,0:31:48.85,Default,,0000,0000,0000,,So definitely take, I would\Nsay, five or six decimal places,
Dialogue: 0,0:31:48.85,0:31:50.35,Default,,0000,0000,0000,,for sure.
Dialogue: 0,0:31:50.35,0:31:54.41,Default,,0000,0000,0000,,So I wrote it as 0.455166.
Dialogue: 0,0:31:54.41,0:31:58.04,Default,,0000,0000,0000,,Now remember, the probability\Nof at least one face card
Dialogue: 0,0:31:58.04,0:32:00.83,Default,,0000,0000,0000,,is 1 minus the probability\Nof no face cards.
Dialogue: 0,0:32:00.83,0:32:03.18,Default,,0000,0000,0000,,And we just calculate the\Nprobability of no face cards.
Dialogue: 0,0:32:03.18,0:32:09.50,Default,,0000,0000,0000,,So 1 minus that\Nvalue is 0.544834.
Dialogue: 0,0:32:09.50,0:32:11.33,Default,,0000,0000,0000,,Now we're done,\Nexcept for the fact
Dialogue: 0,0:32:11.33,0:32:13.67,Default,,0000,0000,0000,,they asked us to do it\Nas a percent rounded
Dialogue: 0,0:32:13.67,0:32:14.72,Default,,0000,0000,0000,,to one decimal place.
Dialogue: 0,0:32:14.72,0:32:17.35,Default,,0000,0000,0000,,So changing it to a percent\Nmoves the decimal place
Dialogue: 0,0:32:17.35,0:32:19.14,Default,,0000,0000,0000,,two places to the right.
Dialogue: 0,0:32:19.14,0:32:22.79,Default,,0000,0000,0000,,And then you can see that that\Nfirst decimal place is a 4.
Dialogue: 0,0:32:22.79,0:32:24.38,Default,,0000,0000,0000,,But the hundreds places is an 8.
Dialogue: 0,0:32:24.38,0:32:26.60,Default,,0000,0000,0000,,So that 4 rounds up to a 5.
Dialogue: 0,0:32:26.60,0:32:30.26,Default,,0000,0000,0000,,So as a percent rounded\Nto one decimal place,
Dialogue: 0,0:32:30.26,0:32:34.94,Default,,0000,0000,0000,,it would be 54.5%.
Dialogue: 0,0:32:34.94,0:32:38.83,Default,,0000,0000,0000,,So if you draw three cards in a\Nstandard deck with replacement,
Dialogue: 0,0:32:38.83,0:32:42.05,Default,,0000,0000,0000,,the probability that at\Nleast one is a face card
Dialogue: 0,0:32:42.05,0:32:45.77,Default,,0000,0000,0000,,is over 50%, 54.% approximately.
Dialogue: 0,0:32:45.77,0:32:48.97,Default,,0000,0000,0000,,
Dialogue: 0,0:32:48.97,0:32:51.52,Default,,0000,0000,0000,,Now in the next section,\Nwe'll learn another method
Dialogue: 0,0:32:51.52,0:32:53.19,Default,,0000,0000,0000,,of solving this type\Nof problem that you
Dialogue: 0,0:32:53.19,0:32:54.46,Default,,0000,0000,0000,,might find a little easier.
Dialogue: 0,0:32:54.46,0:32:58.18,Default,,0000,0000,0000,,But for now, this\Nwill get you there.
Dialogue: 0,0:32:58.18,0:32:59.68,Default,,0000,0000,0000,,Now that I have a\Nproblem like this,
Dialogue: 0,0:32:59.68,0:33:02.62,Default,,0000,0000,0000,,I'm trying to give you\Nvariations of problems that,
Dialogue: 0,0:33:02.62,0:33:04.37,Default,,0000,0000,0000,,on the surface,\Nsound very similar,
Dialogue: 0,0:33:04.37,0:33:06.73,Default,,0000,0000,0000,,but may have a different\Ntechnique of solving them,
Dialogue: 0,0:33:06.73,0:33:10.51,Default,,0000,0000,0000,,because other techniques\Nyou've been using don't apply.
Dialogue: 0,0:33:10.51,0:33:13.07,Default,,0000,0000,0000,,How about this?
Dialogue: 0,0:33:13.07,0:33:15.44,Default,,0000,0000,0000,,A pair of dice is\Nrolled three times.
Dialogue: 0,0:33:15.44,0:33:19.29,Default,,0000,0000,0000,,What is the probability that we\Nget a sum of 6 at least once?
Dialogue: 0,0:33:19.29,0:33:21.77,Default,,0000,0000,0000,,And this time, they're not\Nasking for percent answer,
Dialogue: 0,0:33:21.77,0:33:23.57,Default,,0000,0000,0000,,but they do want the\Nfinal decimal rounded
Dialogue: 0,0:33:23.57,0:33:26.66,Default,,0000,0000,0000,,to three decimal places.
Dialogue: 0,0:33:26.66,0:33:28.91,Default,,0000,0000,0000,,We worked a problem\Nsimilar to this
Dialogue: 0,0:33:28.91,0:33:33.77,Default,,0000,0000,0000,,before, except that we\Ndidn't roll a pair of dice.
Dialogue: 0,0:33:33.77,0:33:35.83,Default,,0000,0000,0000,,We just roll one die.
Dialogue: 0,0:33:35.83,0:33:39.89,Default,,0000,0000,0000,,But I didn't want to\Ngo back and remind you
Dialogue: 0,0:33:39.89,0:33:41.30,Default,,0000,0000,0000,,of all those similarities.
Dialogue: 0,0:33:41.30,0:33:43.37,Default,,0000,0000,0000,,Not only is it\Nsimilar to that one,
Dialogue: 0,0:33:43.37,0:33:46.18,Default,,0000,0000,0000,,but it's also very similar the\None we just worked with cards.
Dialogue: 0,0:33:46.18,0:33:47.60,Default,,0000,0000,0000,,The difference\Nthere is that we're
Dialogue: 0,0:33:47.60,0:33:49.58,Default,,0000,0000,0000,,doing probabilities\Nbased on two dice
Dialogue: 0,0:33:49.58,0:33:53.07,Default,,0000,0000,0000,,here instead of drawing\Ncards from a deck.
Dialogue: 0,0:33:53.07,0:33:55.88,Default,,0000,0000,0000,,And as I said, it's\Nextremely similar to one
Dialogue: 0,0:33:55.88,0:33:59.39,Default,,0000,0000,0000,,we did somewhat\Nearlier in this lecture
Dialogue: 0,0:33:59.39,0:34:04.46,Default,,0000,0000,0000,,where we were rolling just\Na single die, but otherwise,
Dialogue: 0,0:34:04.46,0:34:05.15,Default,,0000,0000,0000,,very similar.
Dialogue: 0,0:34:05.15,0:34:09.61,Default,,0000,0000,0000,,So I put up in the\Ncorner a screen capture
Dialogue: 0,0:34:09.61,0:34:11.53,Default,,0000,0000,0000,,from that calculation.
Dialogue: 0,0:34:11.53,0:34:12.70,Default,,0000,0000,0000,,And you can see what we did.
Dialogue: 0,0:34:12.70,0:34:14.27,Default,,0000,0000,0000,,And we just did it\Na few minutes ago.
Dialogue: 0,0:34:14.27,0:34:16.75,Default,,0000,0000,0000,,So it should be\Nfresh in your mind.
Dialogue: 0,0:34:16.75,0:34:19.54,Default,,0000,0000,0000,,The point being, neither\Nthe problems involving
Dialogue: 0,0:34:19.54,0:34:21.97,Default,,0000,0000,0000,,die rolls, either this one\Nwith two dice or the other one
Dialogue: 0,0:34:21.97,0:34:25.31,Default,,0000,0000,0000,,involving one die, which are\Nindependent events-- rolling
Dialogue: 0,0:34:25.31,0:34:28.48,Default,,0000,0000,0000,,a pair of dice once\Nand rolling them again,
Dialogue: 0,0:34:28.48,0:34:30.98,Default,,0000,0000,0000,,those events are completely\Nindependent of each other.
Dialogue: 0,0:34:30.98,0:34:35.05,Default,,0000,0000,0000,,The probabilities don't change\Nbecause of a previous roll--
Dialogue: 0,0:34:35.05,0:34:38.68,Default,,0000,0000,0000,,nor the previous card problem\Nwhere we drew with replacement.
Dialogue: 0,0:34:38.68,0:34:40.48,Default,,0000,0000,0000,,Because we were drawing\Nwith replacement,
Dialogue: 0,0:34:40.48,0:34:43.36,Default,,0000,0000,0000,,that made those\Ndraws independent.
Dialogue: 0,0:34:43.36,0:34:46.00,Default,,0000,0000,0000,,None of the problems\Ninvolve counting things
Dialogue: 0,0:34:46.00,0:34:48.82,Default,,0000,0000,0000,,because either permutation\Nor combination formulas.
Dialogue: 0,0:34:48.82,0:34:51.13,Default,,0000,0000,0000,,We can't use those\Nwhen we're drawing
Dialogue: 0,0:34:51.13,0:34:53.83,Default,,0000,0000,0000,,with replacement\Nor any case where
Dialogue: 0,0:34:53.83,0:34:55.34,Default,,0000,0000,0000,,there are independent events.
Dialogue: 0,0:34:55.34,0:34:57.40,Default,,0000,0000,0000,,So again, we've got to\Ngo back to the counting
Dialogue: 0,0:34:57.40,0:34:59.45,Default,,0000,0000,0000,,principle for this solution.
Dialogue: 0,0:34:59.45,0:35:03.83,Default,,0000,0000,0000,,But otherwise, it's very similar\Nto the one die roll problem.
Dialogue: 0,0:35:03.83,0:35:06.19,Default,,0000,0000,0000,,If you look at that in\Nthe lower right corner,
Dialogue: 0,0:35:06.19,0:35:10.18,Default,,0000,0000,0000,,the only difference is instead\Nof looking at probabilities
Dialogue: 0,0:35:10.18,0:35:13.21,Default,,0000,0000,0000,,involving getting fours, which\Nis what the other problems are.
Dialogue: 0,0:35:13.21,0:35:16.27,Default,,0000,0000,0000,,We're trying not\Nto get a sum of 6.
Dialogue: 0,0:35:16.27,0:35:19.57,Default,,0000,0000,0000,,Now any time you say the\Nsum of rolling two dice,
Dialogue: 0,0:35:19.57,0:35:22.00,Default,,0000,0000,0000,,you've got to\Nremember that we've
Dialogue: 0,0:35:22.00,0:35:23.80,Default,,0000,0000,0000,,got to go back and look\Nat the sample space
Dialogue: 0,0:35:23.80,0:35:24.82,Default,,0000,0000,0000,,for rolling two dice.
Dialogue: 0,0:35:24.82,0:35:26.17,Default,,0000,0000,0000,,There are 36 possibilities.
Dialogue: 0,0:35:26.17,0:35:29.41,Default,,0000,0000,0000,,We've done this multiple times.
Dialogue: 0,0:35:29.41,0:35:32.54,Default,,0000,0000,0000,,But just to emphasize this, in\Nthe lower right hand corner,
Dialogue: 0,0:35:32.54,0:35:35.44,Default,,0000,0000,0000,,we went from looking for\Nthe probability of getting
Dialogue: 0,0:35:35.44,0:35:39.61,Default,,0000,0000,0000,,any fours when we're rolling\Na single die to what we really
Dialogue: 0,0:35:39.61,0:35:42.79,Default,,0000,0000,0000,,want now, which is the\Nprobability involving
Dialogue: 0,0:35:42.79,0:35:44.29,Default,,0000,0000,0000,,sums of 6.
Dialogue: 0,0:35:44.29,0:35:47.02,Default,,0000,0000,0000,,That takes us to the sample\Nspace in the upper right hand
Dialogue: 0,0:35:47.02,0:35:51.73,Default,,0000,0000,0000,,corner, which is a 36-member\Nor 36-element sample space.
Dialogue: 0,0:35:51.73,0:35:54.45,Default,,0000,0000,0000,,
Dialogue: 0,0:35:54.45,0:35:56.27,Default,,0000,0000,0000,,Once we get this,\Nall we have to do
Dialogue: 0,0:35:56.27,0:35:58.35,Default,,0000,0000,0000,,is figure out the value\Nthat goes in the numerator
Dialogue: 0,0:35:58.35,0:35:58.74,Default,,0000,0000,0000,,and denominator.
Dialogue: 0,0:35:58.74,0:36:01.33,Default,,0000,0000,0000,,And I think maybe it's easier\Nto start with the denominator.
Dialogue: 0,0:36:01.33,0:36:04.23,Default,,0000,0000,0000,,So let's look at the\Ndenominator first.
Dialogue: 0,0:36:04.23,0:36:05.91,Default,,0000,0000,0000,,Using the counting\Nprinciple, remember,
Dialogue: 0,0:36:05.91,0:36:07.38,Default,,0000,0000,0000,,I said we'd have to go\Nback to the counting
Dialogue: 0,0:36:07.38,0:36:08.82,Default,,0000,0000,0000,,principle for these\Nproblems where
Dialogue: 0,0:36:08.82,0:36:11.39,Default,,0000,0000,0000,,the events are independent.
Dialogue: 0,0:36:11.39,0:36:14.90,Default,,0000,0000,0000,,We're going to do three\Nroll of a pair of dice.
Dialogue: 0,0:36:14.90,0:36:16.83,Default,,0000,0000,0000,,On the first roll\Nin the denominator
Dialogue: 0,0:36:16.83,0:36:18.86,Default,,0000,0000,0000,,we don't care what the sum is.
Dialogue: 0,0:36:18.86,0:36:23.23,Default,,0000,0000,0000,,So it could be any\Nof those 36 sums.
Dialogue: 0,0:36:23.23,0:36:25.71,Default,,0000,0000,0000,,So you've got 36 possibilities\Nfor the first row,
Dialogue: 0,0:36:25.71,0:36:28.03,Default,,0000,0000,0000,,36 possibilities\Nfor the second row,
Dialogue: 0,0:36:28.03,0:36:30.75,Default,,0000,0000,0000,,36 possibilities\Nfor the third row.
Dialogue: 0,0:36:30.75,0:36:37.90,Default,,0000,0000,0000,,And 36 times 36 times 36, by the\Ncounting principle, is 46,656.
Dialogue: 0,0:36:37.90,0:36:43.73,Default,,0000,0000,0000,,That's the denominator of our\Nanswer for the probability
Dialogue: 0,0:36:43.73,0:36:47.27,Default,,0000,0000,0000,,of getting no sums of 6.
Dialogue: 0,0:36:47.27,0:36:49.49,Default,,0000,0000,0000,,Let's move to the numerator.
Dialogue: 0,0:36:49.49,0:36:54.48,Default,,0000,0000,0000,,Now the numerator, we\Ndon't want any sums of 6.
Dialogue: 0,0:36:54.48,0:36:55.85,Default,,0000,0000,0000,,We're looking for\Nthe probability
Dialogue: 0,0:36:55.85,0:36:59.57,Default,,0000,0000,0000,,of getting no sums of 6.
Dialogue: 0,0:36:59.57,0:37:04.51,Default,,0000,0000,0000,,Well, if you look up there,\Nthere are 31 of them.
Dialogue: 0,0:37:04.51,0:37:05.77,Default,,0000,0000,0000,,5, 1 adds to 6.
Dialogue: 0,0:37:05.77,0:37:07.42,Default,,0000,0000,0000,,4, 2, adds to 6.
Dialogue: 0,0:37:07.42,0:37:08.83,Default,,0000,0000,0000,,3, 3 adds to 6.
Dialogue: 0,0:37:08.83,0:37:10.57,Default,,0000,0000,0000,,2, 4 adds to 6.
Dialogue: 0,0:37:10.57,0:37:11.74,Default,,0000,0000,0000,,1, 5 adds to 6.
Dialogue: 0,0:37:11.74,0:37:15.07,Default,,0000,0000,0000,,In other words, only\Nthat diagonal line
Dialogue: 0,0:37:15.07,0:37:18.43,Default,,0000,0000,0000,,of sums that I did\Nnot circle add to 6.
Dialogue: 0,0:37:18.43,0:37:20.82,Default,,0000,0000,0000,,We want the ones\Nthat don't add to 6.
Dialogue: 0,0:37:20.82,0:37:24.19,Default,,0000,0000,0000,,And if you count those\Ncircles or just say 36 minus 5
Dialogue: 0,0:37:24.19,0:37:26.14,Default,,0000,0000,0000,,that weren't circled,\Nyou'll see that there
Dialogue: 0,0:37:26.14,0:37:30.18,Default,,0000,0000,0000,,are 31 possibilities.
Dialogue: 0,0:37:30.18,0:37:32.82,Default,,0000,0000,0000,,And again, this is a\Ncounting principal problem
Dialogue: 0,0:37:32.82,0:37:34.35,Default,,0000,0000,0000,,and we've got three rolls.
Dialogue: 0,0:37:34.35,0:37:38.13,Default,,0000,0000,0000,,On the first roll, any of\Nthose 31 that I circled are OK.
Dialogue: 0,0:37:38.13,0:37:41.31,Default,,0000,0000,0000,,Yet none of those\N31 give us sum of 6.
Dialogue: 0,0:37:41.31,0:37:43.41,Default,,0000,0000,0000,,Again, you roll a second try.
Dialogue: 0,0:37:43.41,0:37:44.13,Default,,0000,0000,0000,,Same thing.
Dialogue: 0,0:37:44.13,0:37:45.63,Default,,0000,0000,0000,,31 possibilities.
Dialogue: 0,0:37:45.63,0:37:48.21,Default,,0000,0000,0000,,Third roll again, there\Nare 31 possibilities.
Dialogue: 0,0:37:48.21,0:37:52.74,Default,,0000,0000,0000,,All you're screening for are\Nsums that are not equal to 6.
Dialogue: 0,0:37:52.74,0:37:55.65,Default,,0000,0000,0000,,And by the counting\Nprinciple, 31 times 31 times
Dialogue: 0,0:37:55.65,0:37:58.20,Default,,0000,0000,0000,,31 is the total number.
Dialogue: 0,0:37:58.20,0:38:01.65,Default,,0000,0000,0000,,It comes out to be 29,791.
Dialogue: 0,0:38:01.65,0:38:04.30,Default,,0000,0000,0000,,And that answer goes\Nin the numerator.
Dialogue: 0,0:38:04.30,0:38:10.47,Default,,0000,0000,0000,,So if you take your calculator,\Nyou get about 0.63852.
Dialogue: 0,0:38:10.47,0:38:12.69,Default,,0000,0000,0000,,Again, take plenty more\Ndecimals than you need,
Dialogue: 0,0:38:12.69,0:38:15.09,Default,,0000,0000,0000,,because you want to make\Nsure you're rounding error
Dialogue: 0,0:38:15.09,0:38:18.63,Default,,0000,0000,0000,,doesn't cause you\Nto miss the problem.
Dialogue: 0,0:38:18.63,0:38:22.92,Default,,0000,0000,0000,,Also remember, we're not\Nlooking for the answer, not
Dialogue: 0,0:38:22.92,0:38:26.19,Default,,0000,0000,0000,,the final answer, as\Nbeing the probability
Dialogue: 0,0:38:26.19,0:38:27.66,Default,,0000,0000,0000,,that there are no sums of 6.
Dialogue: 0,0:38:27.66,0:38:29.49,Default,,0000,0000,0000,,We're actually looking\Nfor the probability
Dialogue: 0,0:38:29.49,0:38:31.50,Default,,0000,0000,0000,,that you get at\Nleast one sum of 6.
Dialogue: 0,0:38:31.50,0:38:33.06,Default,,0000,0000,0000,,But we know by the\Ncomplement rule
Dialogue: 0,0:38:33.06,0:38:35.64,Default,,0000,0000,0000,,that the probability of\Nat least one sum of 6
Dialogue: 0,0:38:35.64,0:38:39.36,Default,,0000,0000,0000,,will be 1 minus the\Nprobability of no sums of 6.
Dialogue: 0,0:38:39.36,0:38:42.57,Default,,0000,0000,0000,,We just calculated the\Nprobability of no sums of 6
Dialogue: 0,0:38:42.57,0:38:45.60,Default,,0000,0000,0000,,being the 0.63852.
Dialogue: 0,0:38:45.60,0:38:49.62,Default,,0000,0000,0000,,So if plug that in and\Nsubtract 1 minus that,
Dialogue: 0,0:38:49.62,0:38:53.37,Default,,0000,0000,0000,,you get zero 0.36148.
Dialogue: 0,0:38:53.37,0:38:55.95,Default,,0000,0000,0000,,They ask us not to change\Nit to a percent this time,
Dialogue: 0,0:38:55.95,0:38:58.86,Default,,0000,0000,0000,,but just to round it to\Nthree decimal places.
Dialogue: 0,0:38:58.86,0:39:04.38,Default,,0000,0000,0000,,That would be 0.361 rounded\Nto one decimal place.
Dialogue: 0,0:39:04.38,0:39:08.65,Default,,0000,0000,0000,,So the probability of rolling\Na pair of dice three times
Dialogue: 0,0:39:08.65,0:39:14.92,Default,,0000,0000,0000,,and getting at least\None sum of 6 is 0.361.
Dialogue: 0,0:39:14.92,0:39:18.18,Default,,0000,0000,0000,,In wrapping up this lecture,\NI do want to say one thing.
Dialogue: 0,0:39:18.18,0:39:19.60,Default,,0000,0000,0000,,When you're\Npracticing and working
Dialogue: 0,0:39:19.60,0:39:21.46,Default,,0000,0000,0000,,through these problems\Nand similar problems,
Dialogue: 0,0:39:21.46,0:39:22.90,Default,,0000,0000,0000,,try looking at the big picture.
Dialogue: 0,0:39:22.90,0:39:25.51,Default,,0000,0000,0000,,If you're isolating each\Nproblem as a separate thing
Dialogue: 0,0:39:25.51,0:39:28.66,Default,,0000,0000,0000,,and just concentrating on\Nsolving that problem as you
Dialogue: 0,0:39:28.66,0:39:30.82,Default,,0000,0000,0000,,move from problem to\Nproblem, you're probably
Dialogue: 0,0:39:30.82,0:39:32.07,Default,,0000,0000,0000,,not seeing the big picture.
Dialogue: 0,0:39:32.07,0:39:35.77,Default,,0000,0000,0000,,And when you come back to those\Nkinds of problems on the test,
Dialogue: 0,0:39:35.77,0:39:38.88,Default,,0000,0000,0000,,you may have trouble\Nsolving them.
Dialogue: 0,0:39:38.88,0:39:40.86,Default,,0000,0000,0000,,Now I'm going to\Nmention a few things.
Dialogue: 0,0:39:40.86,0:39:42.64,Default,,0000,0000,0000,,But this is not an\Nexhaustive list.
Dialogue: 0,0:39:42.64,0:39:45.60,Default,,0000,0000,0000,,But for one thing I mean by\Nthat just the stuff that we've
Dialogue: 0,0:39:45.60,0:39:48.97,Default,,0000,0000,0000,,been dealing with recently about\Nwhether you're drawing cards
Dialogue: 0,0:39:48.97,0:39:51.18,Default,,0000,0000,0000,,with replacement or without\Nreplace replacement, when
Dialogue: 0,0:39:51.18,0:39:53.16,Default,,0000,0000,0000,,you're rolling die--
Dialogue: 0,0:39:53.16,0:39:55.62,Default,,0000,0000,0000,,one die, two dice--
Dialogue: 0,0:39:55.62,0:39:57.58,Default,,0000,0000,0000,,you've got to be careful.
Dialogue: 0,0:39:57.58,0:40:00.21,Default,,0000,0000,0000,,If the things involve\Nindependent events
Dialogue: 0,0:40:00.21,0:40:03.78,Default,,0000,0000,0000,,like rolling dice or drawing\Ncards with replacement,
Dialogue: 0,0:40:03.78,0:40:06.02,Default,,0000,0000,0000,,you're generally going back\Nto the counting principle
Dialogue: 0,0:40:06.02,0:40:10.53,Default,,0000,0000,0000,,because the combination and\Npermutation formulas are
Dialogue: 0,0:40:10.53,0:40:13.02,Default,,0000,0000,0000,,generally used to\Ncount arrangements
Dialogue: 0,0:40:13.02,0:40:16.62,Default,,0000,0000,0000,,or choosing objects\Nfrom distinct objects.
Dialogue: 0,0:40:16.62,0:40:18.60,Default,,0000,0000,0000,,And that simply\Ndoesn't imply when
Dialogue: 0,0:40:18.60,0:40:21.87,Default,,0000,0000,0000,,you're drawing with\Nreplacement or rolling dice.
Dialogue: 0,0:40:21.87,0:40:22.83,Default,,0000,0000,0000,,That's the first thing.
Dialogue: 0,0:40:22.83,0:40:25.35,Default,,0000,0000,0000,,You need to notice when\Nyou're reading the problem is
Dialogue: 0,0:40:25.35,0:40:27.48,Default,,0000,0000,0000,,it an independent event--
Dialogue: 0,0:40:27.48,0:40:29.61,Default,,0000,0000,0000,,rolling dice, drawing\Nwith replacement.
Dialogue: 0,0:40:29.61,0:40:30.95,Default,,0000,0000,0000,,Or is it a dependent event?
Dialogue: 0,0:40:30.95,0:40:33.60,Default,,0000,0000,0000,,If it's a dependent event,\Nwhen you're drawing cards
Dialogue: 0,0:40:33.60,0:40:35.52,Default,,0000,0000,0000,,from a deck and not\Nputting them back,
Dialogue: 0,0:40:35.52,0:40:38.16,Default,,0000,0000,0000,,generally you'll be able to\Ncount using your combination
Dialogue: 0,0:40:38.16,0:40:38.82,Default,,0000,0000,0000,,formulas.
Dialogue: 0,0:40:38.82,0:40:41.89,Default,,0000,0000,0000,,So that's one big picture item.
Dialogue: 0,0:40:41.89,0:40:43.89,Default,,0000,0000,0000,,Another thing I've\Nmentioned in closing
Dialogue: 0,0:40:43.89,0:40:46.59,Default,,0000,0000,0000,,is that the usual\Nstrategy when you computed
Dialogue: 0,0:40:46.59,0:40:49.41,Default,,0000,0000,0000,,the probability of at\Nleast one of something,
Dialogue: 0,0:40:49.41,0:40:51.73,Default,,0000,0000,0000,,you handle that by applying\Nthe complement rule.
Dialogue: 0,0:40:51.73,0:40:53.31,Default,,0000,0000,0000,,In other words, go\Nahead and calculate
Dialogue: 0,0:40:53.31,0:40:55.96,Default,,0000,0000,0000,,the probability of getting\Nnone of those things.
Dialogue: 0,0:40:55.96,0:40:57.84,Default,,0000,0000,0000,,And then the probability\Nat least one of them
Dialogue: 0,0:40:57.84,0:41:00.48,Default,,0000,0000,0000,,is 1 minus the probability\Nof none of them.
Dialogue: 0,0:41:00.48,0:41:02.16,Default,,0000,0000,0000,,There are other things\NI could mention,
Dialogue: 0,0:41:02.16,0:41:04.35,Default,,0000,0000,0000,,but I just want to\Nget you thinking
Dialogue: 0,0:41:04.35,0:41:06.73,Default,,0000,0000,0000,,about looking for\Nthe big picture
Dialogue: 0,0:41:06.73,0:41:08.19,Default,,0000,0000,0000,,as you work through\Nthese problems.
Dialogue: 0,0:41:08.19,0:41:12.28,Default,,0000,0000,0000,,It's going to make your test\Ntaking experience much better.
Dialogue: 0,0:41:12.28,0:41:15.09,Default,,0000,0000,0000,,Section 12.5 on\Nconditional probability--
Dialogue: 0,0:41:15.09,0:41:17.51,Default,,0000,0000,0000,,I want to talk about the idea\Nof independent and dependent
Dialogue: 0,0:41:17.51,0:41:18.35,Default,,0000,0000,0000,,events.
Dialogue: 0,0:41:18.35,0:41:20.22,Default,,0000,0000,0000,,I alluded to that in\Nthe earlier formulas
Dialogue: 0,0:41:20.22,0:41:22.83,Default,,0000,0000,0000,,when I put up those\Nformulas in advance.
Dialogue: 0,0:41:22.83,0:41:25.70,Default,,0000,0000,0000,,Independent events are ones for\Nwhich the result of one event
Dialogue: 0,0:41:25.70,0:41:28.32,Default,,0000,0000,0000,,does not affect the probability\Nof the occurrence of the other.
Dialogue: 0,0:41:28.32,0:41:30.53,Default,,0000,0000,0000,,There are some simple examples\Nthat we'll go through.
Dialogue: 0,0:41:30.53,0:41:32.67,Default,,0000,0000,0000,,But I need to throw up a\Nthing to contrast it with,
Dialogue: 0,0:41:32.67,0:41:34.39,Default,,0000,0000,0000,,which are dependent events.
Dialogue: 0,0:41:34.39,0:41:36.16,Default,,0000,0000,0000,,Dependent events\Nare one for which
Dialogue: 0,0:41:36.16,0:41:38.28,Default,,0000,0000,0000,,the occurrence of one event\Naffects the probability
Dialogue: 0,0:41:38.28,0:41:39.98,Default,,0000,0000,0000,,of the occurrence of the other.
Dialogue: 0,0:41:39.98,0:41:41.40,Default,,0000,0000,0000,,Now that you can\Nsee the contrast,
Dialogue: 0,0:41:41.40,0:41:44.28,Default,,0000,0000,0000,,I want I put some\Nexamples up and see
Dialogue: 0,0:41:44.28,0:41:47.10,Default,,0000,0000,0000,,if I can make it even clearer.
Dialogue: 0,0:41:47.10,0:41:48.81,Default,,0000,0000,0000,,Let's decide whether\Nthe following events
Dialogue: 0,0:41:48.81,0:41:51.13,Default,,0000,0000,0000,,are dependent or independent.
Dialogue: 0,0:41:51.13,0:41:53.89,Default,,0000,0000,0000,,Doing your assignment and\Npassing your math class.
Dialogue: 0,0:41:53.89,0:41:55.84,Default,,0000,0000,0000,,Do you think doing\Nyour assignments might
Dialogue: 0,0:41:55.84,0:41:57.82,Default,,0000,0000,0000,,have some effect on\Nthe probability of you
Dialogue: 0,0:41:57.82,0:42:00.58,Default,,0000,0000,0000,,passing your math class?
Dialogue: 0,0:42:00.58,0:42:03.07,Default,,0000,0000,0000,,Yeah, I think most people\Nwould agree to that.
Dialogue: 0,0:42:03.07,0:42:05.62,Default,,0000,0000,0000,,So those are dependent events.
Dialogue: 0,0:42:05.62,0:42:10.32,Default,,0000,0000,0000,,One thing can affect the\Nprobability of the other thing.
Dialogue: 0,0:42:10.32,0:42:11.31,Default,,0000,0000,0000,,How about this?
Dialogue: 0,0:42:11.31,0:42:14.91,Default,,0000,0000,0000,,Running daily and winning\Nthe New York Marathon?
Dialogue: 0,0:42:14.91,0:42:18.33,Default,,0000,0000,0000,,Again, I think most\Npeople would agree
Dialogue: 0,0:42:18.33,0:42:22.74,Default,,0000,0000,0000,,that that might have an effect,\Nthat daily running might very
Dialogue: 0,0:42:22.74,0:42:25.05,Default,,0000,0000,0000,,well have an effect\Non how probably it
Dialogue: 0,0:42:25.05,0:42:27.54,Default,,0000,0000,0000,,is that that person will\Nwin the New York Marathon.
Dialogue: 0,0:42:27.54,0:42:31.26,Default,,0000,0000,0000,,So I'd say again, those\Nare dependent events.
Dialogue: 0,0:42:31.26,0:42:31.88,Default,,0000,0000,0000,,How about this?
Dialogue: 0,0:42:31.88,0:42:35.72,Default,,0000,0000,0000,,Winning the lottery and\Nwinning a spelling bee?
Dialogue: 0,0:42:35.72,0:42:38.12,Default,,0000,0000,0000,,It's really hard to imagine\Nhow winning a lottery
Dialogue: 0,0:42:38.12,0:42:42.26,Default,,0000,0000,0000,,could affect the probability\Nof winning a spelling bee,
Dialogue: 0,0:42:42.26,0:42:44.87,Default,,0000,0000,0000,,or the probability of\Nwinning a spelling bee
Dialogue: 0,0:42:44.87,0:42:47.39,Default,,0000,0000,0000,,would have any effect\Non the probability
Dialogue: 0,0:42:47.39,0:42:48.30,Default,,0000,0000,0000,,of winning a lottery.
Dialogue: 0,0:42:48.30,0:42:51.03,Default,,0000,0000,0000,,
Dialogue: 0,0:42:51.03,0:42:53.27,Default,,0000,0000,0000,,So those two events\Nare independent.
Dialogue: 0,0:42:53.27,0:42:56.16,Default,,0000,0000,0000,,
Dialogue: 0,0:42:56.16,0:42:59.05,Default,,0000,0000,0000,,Getting heads on a coin toss\Nand rolling a 6 on the standard
Dialogue: 0,0:42:59.05,0:43:01.54,Default,,0000,0000,0000,,die.
Dialogue: 0,0:43:01.54,0:43:02.29,Default,,0000,0000,0000,,Flip a coin.
Dialogue: 0,0:43:02.29,0:43:04.39,Default,,0000,0000,0000,,Does it matter what\Nyou get in terms
Dialogue: 0,0:43:04.39,0:43:07.36,Default,,0000,0000,0000,,of it changing the probability\Nyou're going to roll a 6
Dialogue: 0,0:43:07.36,0:43:09.01,Default,,0000,0000,0000,,on the standard die?
Dialogue: 0,0:43:09.01,0:43:10.36,Default,,0000,0000,0000,,No, I don't think so.
Dialogue: 0,0:43:10.36,0:43:12.91,Default,,0000,0000,0000,,So those are also\Nindependent events.
Dialogue: 0,0:43:12.91,0:43:14.38,Default,,0000,0000,0000,,This sort of gives\Nyou an example
Dialogue: 0,0:43:14.38,0:43:16.33,Default,,0000,0000,0000,,of what we mean by\Ndependent and independent.
Dialogue: 0,0:43:16.33,0:43:20.99,Default,,0000,0000,0000,,You may only need an intuitive\Nidea of what they mean.
Dialogue: 0,0:43:20.99,0:43:22.00,Default,,0000,0000,0000,,How about this?
Dialogue: 0,0:43:22.00,0:43:24.58,Default,,0000,0000,0000,,Drawing an ace, then drawing\Na 2 from a deck of cards
Dialogue: 0,0:43:24.58,0:43:27.40,Default,,0000,0000,0000,,without replacement?
Dialogue: 0,0:43:27.40,0:43:28.65,Default,,0000,0000,0000,,Now think about this.
Dialogue: 0,0:43:28.65,0:43:32.76,Default,,0000,0000,0000,,If you draw an ace out,\Nyou don't replace it,
Dialogue: 0,0:43:32.76,0:43:37.29,Default,,0000,0000,0000,,and then you draw a 2, that\Ndoes affect the probability,
Dialogue: 0,0:43:37.29,0:43:42.33,Default,,0000,0000,0000,,because you since you\Ndidn't put that ace back in,
Dialogue: 0,0:43:42.33,0:43:47.25,Default,,0000,0000,0000,,you're only drawing from\N51 cards the second time.
Dialogue: 0,0:43:47.25,0:43:49.14,Default,,0000,0000,0000,,There is a dependency there.
Dialogue: 0,0:43:49.14,0:43:52.23,Default,,0000,0000,0000,,Drawing that ace out and\Nnot putting it back in
Dialogue: 0,0:43:52.23,0:43:55.48,Default,,0000,0000,0000,,makes those two\Nevents dependent.
Dialogue: 0,0:43:55.48,0:43:59.44,Default,,0000,0000,0000,,If you put the ace back\Nin before you drew again,
Dialogue: 0,0:43:59.44,0:44:01.59,Default,,0000,0000,0000,,then that would\Nerase the dependents.
Dialogue: 0,0:44:01.59,0:44:03.56,Default,,0000,0000,0000,,And they would be\Nindependent events.
Dialogue: 0,0:44:03.56,0:44:05.38,Default,,0000,0000,0000,,These are very\Nimportant distinctions.
Dialogue: 0,0:44:05.38,0:44:08.02,Default,,0000,0000,0000,,You need basically\Nan intuitive idea.
Dialogue: 0,0:44:08.02,0:44:11.44,Default,,0000,0000,0000,,But having an intuitive\Nidea is extremely important.
Dialogue: 0,0:44:11.44,0:44:13.90,Default,,0000,0000,0000,,
Dialogue: 0,0:44:13.90,0:44:15.82,Default,,0000,0000,0000,,And this leads to something\Ncalled the product
Dialogue: 0,0:44:15.82,0:44:17.77,Default,,0000,0000,0000,,rule for independent events.
Dialogue: 0,0:44:17.77,0:44:19.94,Default,,0000,0000,0000,,If two events E and\NF are independent,
Dialogue: 0,0:44:19.94,0:44:23.77,Default,,0000,0000,0000,,then the probability of E\Nand F is the probability of E
Dialogue: 0,0:44:23.77,0:44:27.40,Default,,0000,0000,0000,,times the probability of F.\NThat was one of the tools
Dialogue: 0,0:44:27.40,0:44:29.11,Default,,0000,0000,0000,,that I flashed up at\Nthe very beginning,
Dialogue: 0,0:44:29.11,0:44:31.46,Default,,0000,0000,0000,,and said we'd get to it later.
Dialogue: 0,0:44:31.46,0:44:32.74,Default,,0000,0000,0000,,Suppose you flip two coins.
Dialogue: 0,0:44:32.74,0:44:35.14,Default,,0000,0000,0000,,What is the probability\Nthat the first clip is heads
Dialogue: 0,0:44:35.14,0:44:36.70,Default,,0000,0000,0000,,and the second flip is tail?
Dialogue: 0,0:44:36.70,0:44:39.10,Default,,0000,0000,0000,,The coin flips are independent.
Dialogue: 0,0:44:39.10,0:44:42.58,Default,,0000,0000,0000,,What I get the first time I flip\Nthe coin has absolutely nothing
Dialogue: 0,0:44:42.58,0:44:45.22,Default,,0000,0000,0000,,to do with what I get the\Nsecond time I flip the coin.
Dialogue: 0,0:44:45.22,0:44:47.02,Default,,0000,0000,0000,,So that means\Nthey're independent.
Dialogue: 0,0:44:47.02,0:44:48.73,Default,,0000,0000,0000,,So that tells me\Nthat the probability
Dialogue: 0,0:44:48.73,0:44:51.37,Default,,0000,0000,0000,,I get a head and\Nthen a tail is simply
Dialogue: 0,0:44:51.37,0:44:54.55,Default,,0000,0000,0000,,the probability I get ahead\Ntimes the probability I get
Dialogue: 0,0:44:54.55,0:44:57.09,Default,,0000,0000,0000,,a tail, because that's\Nwhat the product rule
Dialogue: 0,0:44:57.09,0:44:58.98,Default,,0000,0000,0000,,for independent events says.
Dialogue: 0,0:44:58.98,0:45:01.18,Default,,0000,0000,0000,,If the probably I\Nget ahead is 1/2.
Dialogue: 0,0:45:01.18,0:45:03.07,Default,,0000,0000,0000,,The probability I\Nget a tail is 1/2.
Dialogue: 0,0:45:03.07,0:45:06.07,Default,,0000,0000,0000,,So the probability that I\Nget a head followed by a tail
Dialogue: 0,0:45:06.07,0:45:08.62,Default,,0000,0000,0000,,is 1/4, a 1/2 times 1/2.
Dialogue: 0,0:45:08.62,0:45:10.58,Default,,0000,0000,0000,,That's the product rule\Nfor independent events.
Dialogue: 0,0:45:10.58,0:45:13.14,Default,,0000,0000,0000,,
Dialogue: 0,0:45:13.14,0:45:15.14,Default,,0000,0000,0000,,I just want to say that\Nyou could have worked
Dialogue: 0,0:45:15.14,0:45:19.37,Default,,0000,0000,0000,,this problem with a tree diagram\Nif you look at your first toss
Dialogue: 0,0:45:19.37,0:45:23.24,Default,,0000,0000,0000,,as being the first stage\Nof a two-stage experiment--
Dialogue: 0,0:45:23.24,0:45:25.52,Default,,0000,0000,0000,,the probability of\Ngetting a head is 1/2.
Dialogue: 0,0:45:25.52,0:45:27.10,Default,,0000,0000,0000,,And then once\Nyou've done a head,
Dialogue: 0,0:45:27.10,0:45:29.66,Default,,0000,0000,0000,,the probability of getting\Nanother head is 1/2,
Dialogue: 0,0:45:29.66,0:45:31.88,Default,,0000,0000,0000,,and the probability of\Ngetting a tail is also 1/2.
Dialogue: 0,0:45:31.88,0:45:35.60,Default,,0000,0000,0000,,So if you look at the branch\Nthat goes from heads to tails,
Dialogue: 0,0:45:35.60,0:45:37.99,Default,,0000,0000,0000,,you see it's 1/2\Nprobability you get ahead,
Dialogue: 0,0:45:37.99,0:45:39.95,Default,,0000,0000,0000,,and there's one 1/2\Nprobability you get a tail.
Dialogue: 0,0:45:39.95,0:45:44.27,Default,,0000,0000,0000,,And we learned earlier that the\Nprobability multiplying down
Dialogue: 0,0:45:44.27,0:45:47.72,Default,,0000,0000,0000,,a branch is the and\Nprobability. so the probability
Dialogue: 0,0:45:47.72,0:45:50.18,Default,,0000,0000,0000,,of getting a head and a\Ntail, where the first is
Dialogue: 0,0:45:50.18,0:45:52.73,Default,,0000,0000,0000,,a head and the second is\Ntail, it would be 1/2 times
Dialogue: 0,0:45:52.73,0:45:55.58,Default,,0000,0000,0000,,1/2, which is 1/4.
Dialogue: 0,0:45:55.58,0:45:57.74,Default,,0000,0000,0000,,So we did learn something\Nin earlier sections
Dialogue: 0,0:45:57.74,0:46:01.10,Default,,0000,0000,0000,,that would allow you to get\Nthat same answer with a tree
Dialogue: 0,0:46:01.10,0:46:02.10,Default,,0000,0000,0000,,diagram.
Dialogue: 0,0:46:02.10,0:46:05.00,Default,,0000,0000,0000,,But treating it as just\Ntwo independent events
Dialogue: 0,0:46:05.00,0:46:07.04,Default,,0000,0000,0000,,is much simpler.
Dialogue: 0,0:46:07.04,0:46:08.93,Default,,0000,0000,0000,,Now I want to talk about\Nconditional events.
Dialogue: 0,0:46:08.93,0:46:12.13,Default,,0000,0000,0000,,That is the subject\Nof section 12.5.
Dialogue: 0,0:46:12.13,0:46:13.98,Default,,0000,0000,0000,,And this is also\Nwhere I said earlier,
Dialogue: 0,0:46:13.98,0:46:17.35,Default,,0000,0000,0000,,we introduced a new notation.
Dialogue: 0,0:46:17.35,0:46:20.95,Default,,0000,0000,0000,,The conditional probability\Nof event E occurring
Dialogue: 0,0:46:20.95,0:46:25.83,Default,,0000,0000,0000,,given that event F has\Noccurred is the probability
Dialogue: 0,0:46:25.83,0:46:28.62,Default,,0000,0000,0000,,of the event E occurring,\Ntaking into account
Dialogue: 0,0:46:28.62,0:46:30.52,Default,,0000,0000,0000,,that event F has\Nalready occurred.
Dialogue: 0,0:46:30.52,0:46:34.96,Default,,0000,0000,0000,,And that's what the\Nadjective conditional means.
Dialogue: 0,0:46:34.96,0:46:38.47,Default,,0000,0000,0000,,Conditional probability\Nmeans the probability
Dialogue: 0,0:46:38.47,0:46:41.41,Default,,0000,0000,0000,,conditioned on the fact that\Nsomething else has already
Dialogue: 0,0:46:41.41,0:46:42.42,Default,,0000,0000,0000,,occurred.
Dialogue: 0,0:46:42.42,0:46:45.32,Default,,0000,0000,0000,,It's good to probably introduce\Nthis idea through an example.
Dialogue: 0,0:46:45.32,0:46:47.29,Default,,0000,0000,0000,,Let's let E be the\Nevent that the sum is
Dialogue: 0,0:46:47.29,0:46:50.45,Default,,0000,0000,0000,,5 when we're rolling two dice.
Dialogue: 0,0:46:50.45,0:46:52.33,Default,,0000,0000,0000,,What is the probability of E?
Dialogue: 0,0:46:52.33,0:46:54.55,Default,,0000,0000,0000,,Well, that's just the\Nprobability we get a sum of 5.
Dialogue: 0,0:46:54.55,0:46:55.72,Default,,0000,0000,0000,,And we know how to do that.
Dialogue: 0,0:46:55.72,0:46:57.49,Default,,0000,0000,0000,,We draw our 2 by 2 table.
Dialogue: 0,0:46:57.49,0:47:01.63,Default,,0000,0000,0000,,And we count the sums that come\Nto 5, which happened to be 4.
Dialogue: 0,0:47:01.63,0:47:04.62,Default,,0000,0000,0000,,And we get, by the basic\Nprobability principle
Dialogue: 0,0:47:04.62,0:47:07.03,Default,,0000,0000,0000,,is the number of ways of\Ngetting what you're looking for,
Dialogue: 0,0:47:07.03,0:47:10.24,Default,,0000,0000,0000,,which in this case is the sum of\N5 divided by the number of ways
Dialogue: 0,0:47:10.24,0:47:12.22,Default,,0000,0000,0000,,you can get anything in\Nthe sample space, which
Dialogue: 0,0:47:12.22,0:47:13.64,Default,,0000,0000,0000,,we know there 36 things.
Dialogue: 0,0:47:13.64,0:47:15.37,Default,,0000,0000,0000,,So it would be 4 over 36.
Dialogue: 0,0:47:15.37,0:47:17.80,Default,,0000,0000,0000,,If you want to reduce\Nit to 1/9, you can.
Dialogue: 0,0:47:17.80,0:47:20.23,Default,,0000,0000,0000,,But what if I tell you\Nthat I've already peaked?
Dialogue: 0,0:47:20.23,0:47:22.06,Default,,0000,0000,0000,,I've done the die roll already.
Dialogue: 0,0:47:22.06,0:47:23.89,Default,,0000,0000,0000,,And I'm holding my hand\Nso you can't see it.
Dialogue: 0,0:47:23.89,0:47:26.17,Default,,0000,0000,0000,,But I've peaked.
Dialogue: 0,0:47:26.17,0:47:28.39,Default,,0000,0000,0000,,And I'm telling you\Nsomething about it.
Dialogue: 0,0:47:28.39,0:47:30.13,Default,,0000,0000,0000,,I'm not telling you\Nwhat the sum is,
Dialogue: 0,0:47:30.13,0:47:32.82,Default,,0000,0000,0000,,but I'm going to\Ntell you-- well,
Dialogue: 0,0:47:32.82,0:47:36.44,Default,,0000,0000,0000,,I'm not telling you what\Nit is, but it's odd.
Dialogue: 0,0:47:36.44,0:47:37.31,Default,,0000,0000,0000,,Does that matter?
Dialogue: 0,0:47:37.31,0:47:39.26,Default,,0000,0000,0000,,Does that change\Nthe probabilities
Dialogue: 0,0:47:39.26,0:47:43.60,Default,,0000,0000,0000,,because I gave you\Nsome new information?
Dialogue: 0,0:47:43.60,0:47:45.04,Default,,0000,0000,0000,,And the answer is\Nin general, yes,
Dialogue: 0,0:47:45.04,0:47:47.05,Default,,0000,0000,0000,,that will make a difference.
Dialogue: 0,0:47:47.05,0:47:49.67,Default,,0000,0000,0000,,So what I really want\Nnow is the probability
Dialogue: 0,0:47:49.67,0:47:53.56,Default,,0000,0000,0000,,of the event E happens,\Nwhich is getting a sum of 5,
Dialogue: 0,0:47:53.56,0:47:55.51,Default,,0000,0000,0000,,given that additional\Ninformation, given
Dialogue: 0,0:47:55.51,0:47:57.64,Default,,0000,0000,0000,,that the sum is odd.
Dialogue: 0,0:47:57.64,0:47:59.36,Default,,0000,0000,0000,,That additional\Ninformation is important.
Dialogue: 0,0:47:59.36,0:48:01.63,Default,,0000,0000,0000,,Because if I know\Nthe sum is odd,
Dialogue: 0,0:48:01.63,0:48:03.43,Default,,0000,0000,0000,,that changes the\Nsample space, because I
Dialogue: 0,0:48:03.43,0:48:06.01,Default,,0000,0000,0000,,don't have to consider\Nany even sums anymore.
Dialogue: 0,0:48:06.01,0:48:08.41,Default,,0000,0000,0000,,So I can get rid of every\Nsum that comes out even.
Dialogue: 0,0:48:08.41,0:48:09.43,Default,,0000,0000,0000,,1 plus 1 is 2.
Dialogue: 0,0:48:09.43,0:48:10.33,Default,,0000,0000,0000,,That's even.
Dialogue: 0,0:48:10.33,0:48:11.59,Default,,0000,0000,0000,,5 plus 3 is 8.
Dialogue: 0,0:48:11.59,0:48:12.70,Default,,0000,0000,0000,,That's even.
Dialogue: 0,0:48:12.70,0:48:14.03,Default,,0000,0000,0000,,5 plus 5 is 10.
Dialogue: 0,0:48:14.03,0:48:14.53,Default,,0000,0000,0000,,That's even.
Dialogue: 0,0:48:14.53,0:48:16.30,Default,,0000,0000,0000,,6 plus 6 is 12.
Dialogue: 0,0:48:16.30,0:48:16.87,Default,,0000,0000,0000,,That's even.
Dialogue: 0,0:48:16.87,0:48:19.90,Default,,0000,0000,0000,,So I can get rid\Nof all of those.
Dialogue: 0,0:48:19.90,0:48:22.72,Default,,0000,0000,0000,,So that additional information\Nis extremely useful.
Dialogue: 0,0:48:22.72,0:48:27.63,Default,,0000,0000,0000,,I can throw out the things\Nthat no longer contend.
Dialogue: 0,0:48:27.63,0:48:32.66,Default,,0000,0000,0000,,Because I know it's odd, the\Nevens don't content anymore.
Dialogue: 0,0:48:32.66,0:48:35.89,Default,,0000,0000,0000,,So the probability of E\Ngiven that the sum is odd
Dialogue: 0,0:48:35.89,0:48:37.54,Default,,0000,0000,0000,,is just the number\Nof ways that E
Dialogue: 0,0:48:37.54,0:48:39.73,Default,,0000,0000,0000,,can happen with that new\Ncondition-- in this case,
Dialogue: 0,0:48:39.73,0:48:43.72,Default,,0000,0000,0000,,that the sum is odd-- divided\Nby how the samples place changes
Dialogue: 0,0:48:43.72,0:48:45.52,Default,,0000,0000,0000,,given that new condition.
Dialogue: 0,0:48:45.52,0:48:48.11,Default,,0000,0000,0000,,You're still doing a\Ncount of the event E.
Dialogue: 0,0:48:48.11,0:48:50.65,Default,,0000,0000,0000,,And you're still doing\Na count of event S,
Dialogue: 0,0:48:50.65,0:48:55.06,Default,,0000,0000,0000,,but you're taking into\Nconsideration that something
Dialogue: 0,0:48:55.06,0:48:57.24,Default,,0000,0000,0000,,has happened and you\Nhave more information
Dialogue: 0,0:48:57.24,0:48:58.36,Default,,0000,0000,0000,,than you had to start with.
Dialogue: 0,0:48:58.36,0:49:02.07,Default,,0000,0000,0000,,That's what I mean when\NI say with condition.
Dialogue: 0,0:49:02.07,0:49:05.28,Default,,0000,0000,0000,,Well, once I crossed out all\Nthe even sums, there were 36.
Dialogue: 0,0:49:05.28,0:49:06.81,Default,,0000,0000,0000,,Half of the sums were even.
Dialogue: 0,0:49:06.81,0:49:09.27,Default,,0000,0000,0000,,That knocked the sample\Nspace down to 18.
Dialogue: 0,0:49:09.27,0:49:12.72,Default,,0000,0000,0000,,There are only 18 odd sums.
Dialogue: 0,0:49:12.72,0:49:13.92,Default,,0000,0000,0000,,And then I look--
Dialogue: 0,0:49:13.92,0:49:16.90,Default,,0000,0000,0000,,none of these sums adding\Nto 5 got knocked out,
Dialogue: 0,0:49:16.90,0:49:18.00,Default,,0000,0000,0000,,because they were all odd.
Dialogue: 0,0:49:18.00,0:49:20.73,Default,,0000,0000,0000,,So there are still\Nfour of those.
Dialogue: 0,0:49:20.73,0:49:24.18,Default,,0000,0000,0000,,So the probability\Nof the sum being 5,
Dialogue: 0,0:49:24.18,0:49:30.51,Default,,0000,0000,0000,,if I know that the sum was\Nodd, is going to be 4/18.
Dialogue: 0,0:49:30.51,0:49:33.24,Default,,0000,0000,0000,,It would have been 4/26\Nhad I not known that.
Dialogue: 0,0:49:33.24,0:49:36.05,Default,,0000,0000,0000,,
Dialogue: 0,0:49:36.05,0:49:38.96,Default,,0000,0000,0000,,So instead of the\Nprobability of the sum being
Dialogue: 0,0:49:38.96,0:49:42.92,Default,,0000,0000,0000,,5 being 1/9, now the\Nprobability that the sum is
Dialogue: 0,0:49:42.92,0:49:45.98,Default,,0000,0000,0000,,5 with that\Nadditional knowledge--
Dialogue: 0,0:49:45.98,0:49:48.47,Default,,0000,0000,0000,,that knowledge being\Nthat the sum was odd--
Dialogue: 0,0:49:48.47,0:49:51.44,Default,,0000,0000,0000,,now has doubled to 2/9.
Dialogue: 0,0:49:51.44,0:49:52.87,Default,,0000,0000,0000,,Yes, it does matter.
Dialogue: 0,0:49:52.87,0:49:54.20,Default,,0000,0000,0000,,Or at least it can matter.
Dialogue: 0,0:49:54.20,0:49:57.26,Default,,0000,0000,0000,,You can see here the\Nprobability doubled.
Dialogue: 0,0:49:57.26,0:50:00.54,Default,,0000,0000,0000,,Knowing that information\Ndoubled the probability.
Dialogue: 0,0:50:00.54,0:50:03.44,Default,,0000,0000,0000,,So to sum up, when we\Nhave a probability which
Dialogue: 0,0:50:03.44,0:50:05.22,Default,,0000,0000,0000,,includes an extra condition--
Dialogue: 0,0:50:05.22,0:50:08.00,Default,,0000,0000,0000,,and in this problem, it\Nwas that the sum was odd--
Dialogue: 0,0:50:08.00,0:50:10.77,Default,,0000,0000,0000,,that probability is called\Na conditional probability.
Dialogue: 0,0:50:10.77,0:50:13.07,Default,,0000,0000,0000,,And we have a special\Nnotation for it.
Dialogue: 0,0:50:13.07,0:50:15.21,Default,,0000,0000,0000,,And if we go back to\Nthat previous example,
Dialogue: 0,0:50:15.21,0:50:16.94,Default,,0000,0000,0000,,the probability of E--
Dialogue: 0,0:50:16.94,0:50:19.01,Default,,0000,0000,0000,,I just wrote this\Nsort of intuitively--
Dialogue: 0,0:50:19.01,0:50:23.53,Default,,0000,0000,0000,,the probability of E comma\Ngiven that the sum is odd--
Dialogue: 0,0:50:23.53,0:50:26.52,Default,,0000,0000,0000,,we're going to modify that.
Dialogue: 0,0:50:26.52,0:50:28.50,Default,,0000,0000,0000,,I just wrote that out\Nkind of intuitively.
Dialogue: 0,0:50:28.50,0:50:31.41,Default,,0000,0000,0000,,Suppose we call that\Ncondition F. In other words,
Dialogue: 0,0:50:31.41,0:50:34.12,Default,,0000,0000,0000,,the sum being odd is a\Nnew condition called F.
Dialogue: 0,0:50:34.12,0:50:37.89,Default,,0000,0000,0000,,So the sum being 5\Nwas the event E. Now
Dialogue: 0,0:50:37.89,0:50:43.54,Default,,0000,0000,0000,,we're going to call the some\Nsum odd, the condition F,
Dialogue: 0,0:50:43.54,0:50:45.33,Default,,0000,0000,0000,,then we'd write the\Nconditional probability
Dialogue: 0,0:50:45.33,0:50:46.68,Default,,0000,0000,0000,,with this vertical line.
Dialogue: 0,0:50:46.68,0:50:52.20,Default,,0000,0000,0000,,The probability of E\Nand the vertical line
Dialogue: 0,0:50:52.20,0:50:56.07,Default,,0000,0000,0000,,is read as given that.
Dialogue: 0,0:50:56.07,0:51:04.63,Default,,0000,0000,0000,,So that's the probability of E\Ngiven F. So that new notation,
Dialogue: 0,0:51:04.63,0:51:07.03,Default,,0000,0000,0000,,the vertical line is read given.
Dialogue: 0,0:51:07.03,0:51:11.29,Default,,0000,0000,0000,,So this is the probability\Nof E given F, or given
Dialogue: 0,0:51:11.29,0:51:12.82,Default,,0000,0000,0000,,that F has occurred.
Dialogue: 0,0:51:12.82,0:51:15.65,Default,,0000,0000,0000,,
Dialogue: 0,0:51:15.65,0:51:20.70,Default,,0000,0000,0000,,Probability of E given F.\NThat's conditional probability.
Dialogue: 0,0:51:20.70,0:51:24.06,Default,,0000,0000,0000,,
Dialogue: 0,0:51:24.06,0:51:24.82,Default,,0000,0000,0000,,This is important.
Dialogue: 0,0:51:24.82,0:51:27.06,Default,,0000,0000,0000,,So take a minute to soak it in.
Dialogue: 0,0:51:27.06,0:51:30.06,Default,,0000,0000,0000,,The probability of E given F\Nis a conditional probability
Dialogue: 0,0:51:30.06,0:51:33.00,Default,,0000,0000,0000,,of event E given\Nthat F has occurred,
Dialogue: 0,0:51:33.00,0:51:35.92,Default,,0000,0000,0000,,that F is the additional\Ninformation that we now know
Dialogue: 0,0:51:35.92,0:51:37.05,Default,,0000,0000,0000,,that we didn't know before.
Dialogue: 0,0:51:37.05,0:51:39.92,Default,,0000,0000,0000,,
Dialogue: 0,0:51:39.92,0:51:43.10,Default,,0000,0000,0000,,The vertical line is\Nread given or given that,
Dialogue: 0,0:51:43.10,0:51:47.30,Default,,0000,0000,0000,,depending on exactly how\Nyou want to phrase it.
Dialogue: 0,0:51:47.30,0:51:50.60,Default,,0000,0000,0000,,For our particular problem,\Nthis would be the probability
Dialogue: 0,0:51:50.60,0:51:53.48,Default,,0000,0000,0000,,that the sum is 5 given\Nthat the sum is odd.
Dialogue: 0,0:51:53.48,0:51:56.78,Default,,0000,0000,0000,,
Dialogue: 0,0:51:56.78,0:51:58.82,Default,,0000,0000,0000,,Think about that a\Nminute, because you
Dialogue: 0,0:51:58.82,0:52:04.16,Default,,0000,0000,0000,,need to understand what the\Nprobability is saying or asking
Dialogue: 0,0:52:04.16,0:52:07.27,Default,,0000,0000,0000,,before you can get an answer.
Dialogue: 0,0:52:07.27,0:52:10.11,Default,,0000,0000,0000,,And if I don't make any\Nother point in this lecture,
Dialogue: 0,0:52:10.11,0:52:13.08,Default,,0000,0000,0000,,I want to say this\Nand make this point.
Dialogue: 0,0:52:13.08,0:52:16.31,Default,,0000,0000,0000,,Oftentimes, if you understand\Nhow the given that condition
Dialogue: 0,0:52:16.31,0:52:17.88,Default,,0000,0000,0000,,restricts the sample space--
Dialogue: 0,0:52:17.88,0:52:19.34,Default,,0000,0000,0000,,the original sample\Nspace-- you can
Dialogue: 0,0:52:19.34,0:52:21.35,Default,,0000,0000,0000,,solve this conditional\Nprobability problem
Dialogue: 0,0:52:21.35,0:52:23.15,Default,,0000,0000,0000,,without even formally\Nusing formulas.
Dialogue: 0,0:52:23.15,0:52:27.05,Default,,0000,0000,0000,,
Dialogue: 0,0:52:27.05,0:52:30.95,Default,,0000,0000,0000,,But if you really understand\Nhow the given that condition
Dialogue: 0,0:52:30.95,0:52:33.59,Default,,0000,0000,0000,,restricts the\Noriginal sample space,
Dialogue: 0,0:52:33.59,0:52:36.56,Default,,0000,0000,0000,,quite often you don't\Nreally even need formulas.
Dialogue: 0,0:52:36.56,0:52:39.44,Default,,0000,0000,0000,,For example, if I told you\Nthat two cards are drawn
Dialogue: 0,0:52:39.44,0:52:42.44,Default,,0000,0000,0000,,without replacement from\Nan ordinary deck of cards,
Dialogue: 0,0:52:42.44,0:52:44.72,Default,,0000,0000,0000,,and asked you the probability\Nthat the second card
Dialogue: 0,0:52:44.72,0:52:47.45,Default,,0000,0000,0000,,is a heart, given that the\Nfirst card was a heart--
Dialogue: 0,0:52:47.45,0:52:54.17,Default,,0000,0000,0000,,
Dialogue: 0,0:52:54.17,0:52:55.71,Default,,0000,0000,0000,,They're asking me\Nfor the probability
Dialogue: 0,0:52:55.71,0:52:57.42,Default,,0000,0000,0000,,the second card\Nis a heart, given
Dialogue: 0,0:52:57.42,0:52:59.83,Default,,0000,0000,0000,,that the first was a heart.
Dialogue: 0,0:52:59.83,0:53:01.72,Default,,0000,0000,0000,,Well, if I've noticed\Nthe first was a heart--
Dialogue: 0,0:53:01.72,0:53:05.02,Default,,0000,0000,0000,,I'm given the first\Ncard was a heart, then
Dialogue: 0,0:53:05.02,0:53:06.54,Default,,0000,0000,0000,,that reduces my sample space.
Dialogue: 0,0:53:06.54,0:53:12.44,Default,,0000,0000,0000,,If I know the first card was\Na heart, when I draw again,
Dialogue: 0,0:53:12.44,0:53:15.41,Default,,0000,0000,0000,,instead of having 52\Ncards in the deck,
Dialogue: 0,0:53:15.41,0:53:19.61,Default,,0000,0000,0000,,I'll only have 51\Ncards in the deck.
Dialogue: 0,0:53:19.61,0:53:21.57,Default,,0000,0000,0000,,And since the first\Ncard was a heart
Dialogue: 0,0:53:21.57,0:53:23.91,Default,,0000,0000,0000,,and there are 13\Nhearts in the deck,
Dialogue: 0,0:53:23.91,0:53:27.66,Default,,0000,0000,0000,,if the first card really was\Na heart, after I draw a heart
Dialogue: 0,0:53:27.66,0:53:30.68,Default,,0000,0000,0000,,out, I only have 12 left.
Dialogue: 0,0:53:30.68,0:53:34.66,Default,,0000,0000,0000,,So knowing that the\Nfirst card was a heart,
Dialogue: 0,0:53:34.66,0:53:36.34,Default,,0000,0000,0000,,if I didn't put the\Ncard back, and it
Dialogue: 0,0:53:36.34,0:53:40.24,Default,,0000,0000,0000,,says drawing without\Nreplacement, when I draw again
Dialogue: 0,0:53:40.24,0:53:42.28,Default,,0000,0000,0000,,or just before I\Ndraw again, I'll
Dialogue: 0,0:53:42.28,0:53:46.39,Default,,0000,0000,0000,,be drawing from a 51-card\Ndeck that only has 12 heart.
Dialogue: 0,0:53:46.39,0:53:50.20,Default,,0000,0000,0000,,So now the probability of\Nthat second card being a heart
Dialogue: 0,0:53:50.20,0:53:54.12,Default,,0000,0000,0000,,is simply 12 out of 51.
Dialogue: 0,0:53:54.12,0:53:55.98,Default,,0000,0000,0000,,Knowing that the\Nfirst card was a heart
Dialogue: 0,0:53:55.98,0:53:59.52,Default,,0000,0000,0000,,and it wasn't re-replaced,\Nreduced the sample space down
Dialogue: 0,0:53:59.52,0:54:03.12,Default,,0000,0000,0000,,to 51 cards, 12 of\Nthem being hearts.
Dialogue: 0,0:54:03.12,0:54:08.05,Default,,0000,0000,0000,,And that probability\Nwould be 12 out of 51.
Dialogue: 0,0:54:08.05,0:54:10.93,Default,,0000,0000,0000,,Simple, if you know\Nhow to think about it.
Dialogue: 0,0:54:10.93,0:54:14.25,Default,,0000,0000,0000,,And if you want to reduce\Nthat to 4/17, you can.
Dialogue: 0,0:54:14.25,0:54:15.53,Default,,0000,0000,0000,,How about this one?
Dialogue: 0,0:54:15.53,0:54:19.29,Default,,0000,0000,0000,,A container has 35\Ngreen marbles, 20 blue,
Dialogue: 0,0:54:19.29,0:54:21.27,Default,,0000,0000,0000,,and 4 red marbles.
Dialogue: 0,0:54:21.27,0:54:24.30,Default,,0000,0000,0000,,Two marbles are randomly\Nselected without replacement.
Dialogue: 0,0:54:24.30,0:54:27.36,Default,,0000,0000,0000,,If E is the event that a green\Nmarble is selected first,
Dialogue: 0,0:54:27.36,0:54:30.63,Default,,0000,0000,0000,,and F is the event that the\Nsecond marble is not green,
Dialogue: 0,0:54:30.63,0:54:34.58,Default,,0000,0000,0000,,compute the probability\Nof F given E.
Dialogue: 0,0:54:34.58,0:54:37.90,Default,,0000,0000,0000,,I will say that when I say these\Nproblems where the events are
Dialogue: 0,0:54:37.90,0:54:41.56,Default,,0000,0000,0000,,given letter names, if they tell\Nme what the actual events are,
Dialogue: 0,0:54:41.56,0:54:43.87,Default,,0000,0000,0000,,my inclination is to\Ngo ahead and replace
Dialogue: 0,0:54:43.87,0:54:46.93,Default,,0000,0000,0000,,the letters with an actual\Nphrase representing the event.
Dialogue: 0,0:54:46.93,0:54:50.41,Default,,0000,0000,0000,,So if F is the event that\Nthe second is not green,
Dialogue: 0,0:54:50.41,0:54:51.85,Default,,0000,0000,0000,,I'll just write that out.
Dialogue: 0,0:54:51.85,0:54:55.42,Default,,0000,0000,0000,,And if E is the probability\Nthat the first was green,
Dialogue: 0,0:54:55.42,0:54:57.71,Default,,0000,0000,0000,,I would just write that out.
Dialogue: 0,0:54:57.71,0:54:58.62,Default,,0000,0000,0000,,It's up to you.
Dialogue: 0,0:54:58.62,0:55:01.00,Default,,0000,0000,0000,,But I think it's easier\Nif you do it that way.
Dialogue: 0,0:55:01.00,0:55:03.55,Default,,0000,0000,0000,,
Dialogue: 0,0:55:03.55,0:55:06.69,Default,,0000,0000,0000,,Now having done\Nthat, remember we're
Dialogue: 0,0:55:06.69,0:55:13.74,Default,,0000,0000,0000,,given that the first draw\Nresulted in a green marble.
Dialogue: 0,0:55:13.74,0:55:16.45,Default,,0000,0000,0000,,And we're doing it\Nwithout replacement.
Dialogue: 0,0:55:16.45,0:55:21.17,Default,,0000,0000,0000,,So if there started\Noff being 59 marbles--
Dialogue: 0,0:55:21.17,0:55:24.11,Default,,0000,0000,0000,,after the first draw,\Nthere are only 58 marbles.
Dialogue: 0,0:55:24.11,0:55:26.57,Default,,0000,0000,0000,,And the one that was\Ndrawn was a green.
Dialogue: 0,0:55:26.57,0:55:29.96,Default,,0000,0000,0000,,So instead of 35\Nthere are 34 green.
Dialogue: 0,0:55:29.96,0:55:32.42,Default,,0000,0000,0000,,We want to calculate\Nthe probability
Dialogue: 0,0:55:32.42,0:55:36.94,Default,,0000,0000,0000,,that the second was not green,\Ngiven the first was green.
Dialogue: 0,0:55:36.94,0:55:39.15,Default,,0000,0000,0000,,You can see that I've already\Nput the answer up there
Dialogue: 0,0:55:39.15,0:55:41.18,Default,,0000,0000,0000,,as being 24 over 58.
Dialogue: 0,0:55:41.18,0:55:43.94,Default,,0000,0000,0000,,And I think you can see\Nthat where that came from.
Dialogue: 0,0:55:43.94,0:55:47.12,Default,,0000,0000,0000,,The denominator is\Nsimply the count
Dialogue: 0,0:55:47.12,0:55:50.90,Default,,0000,0000,0000,,of how many marbles were left.
Dialogue: 0,0:55:50.90,0:55:55.29,Default,,0000,0000,0000,,And the numerator are\Nthe number of things
Dialogue: 0,0:55:55.29,0:55:57.60,Default,,0000,0000,0000,,that are not green, the\Nnumber of marbles not green.
Dialogue: 0,0:55:57.60,0:56:00.24,Default,,0000,0000,0000,,And you can see is there\Nare 20 blue and 4 red.
Dialogue: 0,0:56:00.24,0:56:04.35,Default,,0000,0000,0000,,And 20 plus 4 is 24.
Dialogue: 0,0:56:04.35,0:56:08.99,Default,,0000,0000,0000,,So there were 24\Nnon-greens to choose from.
Dialogue: 0,0:56:08.99,0:56:14.44,Default,,0000,0000,0000,,So the probability is\N24 over 58 non-greens
Dialogue: 0,0:56:14.44,0:56:16.39,Default,,0000,0000,0000,,over total marbles left.
Dialogue: 0,0:56:16.39,0:56:18.97,Default,,0000,0000,0000,,And it's 58 instead of\N59, because we're were
Dialogue: 0,0:56:18.97,0:56:21.10,Default,,0000,0000,0000,,drawing without replacement.
Dialogue: 0,0:56:21.10,0:56:24.10,Default,,0000,0000,0000,,If you want to reduce that,\Nyou can reduce it to 12/29,
Dialogue: 0,0:56:24.10,0:56:25.87,Default,,0000,0000,0000,,but Web Assign\Nreally doesn't care.
Dialogue: 0,0:56:25.87,0:56:29.96,Default,,0000,0000,0000,,
Dialogue: 0,0:56:29.96,0:56:31.01,Default,,0000,0000,0000,,How about this one?
Dialogue: 0,0:56:31.01,0:56:32.07,Default,,0000,0000,0000,,Two dice are rolled.
Dialogue: 0,0:56:32.07,0:56:34.29,Default,,0000,0000,0000,,What is the probability\Nthat the sum of the dice
Dialogue: 0,0:56:34.29,0:56:37.26,Default,,0000,0000,0000,,is 6, given that the\Nrow was a double?
Dialogue: 0,0:56:37.26,0:56:41.10,Default,,0000,0000,0000,,Again, you want your sample\Nspace in a table, 2 by 2.
Dialogue: 0,0:56:41.10,0:56:43.02,Default,,0000,0000,0000,,Six possibilities\Nfor the first roll.
Dialogue: 0,0:56:43.02,0:56:45.62,Default,,0000,0000,0000,,Six possibilities\Nfor the second roll.
Dialogue: 0,0:56:45.62,0:56:47.94,Default,,0000,0000,0000,,You're given that the\Nroll was a double.
Dialogue: 0,0:56:47.94,0:56:53.06,Default,,0000,0000,0000,,So we went the sum to be 6\Nwith the additional information
Dialogue: 0,0:56:53.06,0:56:54.79,Default,,0000,0000,0000,,that we rolled a double.
Dialogue: 0,0:56:54.79,0:56:57.50,Default,,0000,0000,0000,,So the probability that\Nthe sum is 6, given
Dialogue: 0,0:56:57.50,0:56:58.93,Default,,0000,0000,0000,,that the roll was a double.
Dialogue: 0,0:56:58.93,0:57:01.34,Default,,0000,0000,0000,,Well, if we know the\Nroll is a double,
Dialogue: 0,0:57:01.34,0:57:02.97,Default,,0000,0000,0000,,we know that's already occurred.
Dialogue: 0,0:57:02.97,0:57:05.00,Default,,0000,0000,0000,,We can get rid of\Neverything but the doubles.
Dialogue: 0,0:57:05.00,0:57:09.56,Default,,0000,0000,0000,,So the doubles are 1, 1; 2,\N2; 3, 3; 4, 4; 5, 5; 6, 6.
Dialogue: 0,0:57:09.56,0:57:13.69,Default,,0000,0000,0000,,So all of a sudden, that\N36-member sample space
Dialogue: 0,0:57:13.69,0:57:18.62,Default,,0000,0000,0000,,reduced down to 6, because\Nwe know that it was a double.
Dialogue: 0,0:57:18.62,0:57:22.27,Default,,0000,0000,0000,,So we'll be dividing by 6.
Dialogue: 0,0:57:22.27,0:57:26.05,Default,,0000,0000,0000,,Out of those\Nremaining 6 choices,
Dialogue: 0,0:57:26.05,0:57:27.67,Default,,0000,0000,0000,,only one has a sum of 6.
Dialogue: 0,0:57:27.67,0:57:28.84,Default,,0000,0000,0000,,And that's 3, 3.
Dialogue: 0,0:57:28.84,0:57:29.85,Default,,0000,0000,0000,,3 plus 3 is 6.
Dialogue: 0,0:57:29.85,0:57:34.08,Default,,0000,0000,0000,,So the probability\Nis 1 out of 6 or 1/6.
Dialogue: 0,0:57:34.08,0:57:34.89,Default,,0000,0000,0000,,See how this works?
Dialogue: 0,0:57:34.89,0:57:37.08,Default,,0000,0000,0000,,If you can think it\Nthrough, you don't really
Dialogue: 0,0:57:37.08,0:57:38.94,Default,,0000,0000,0000,,have to memorize\Nthe definitions.
Dialogue: 0,0:57:38.94,0:57:42.93,Default,,0000,0000,0000,,You just have to understand\Nwhat the condition does
Dialogue: 0,0:57:42.93,0:57:43.86,Default,,0000,0000,0000,,to the sample space.
Dialogue: 0,0:57:43.86,0:57:47.24,Default,,0000,0000,0000,,
Dialogue: 0,0:57:47.24,0:57:48.44,Default,,0000,0000,0000,,How about this one?
Dialogue: 0,0:57:48.44,0:57:50.80,Default,,0000,0000,0000,,Mrs. Fraga's class\Nhas 109 students
Dialogue: 0,0:57:50.80,0:57:53.14,Default,,0000,0000,0000,,classified by year and gender.
Dialogue: 0,0:57:53.14,0:57:54.91,Default,,0000,0000,0000,,It's listed in the table below.
Dialogue: 0,0:57:54.91,0:57:58.15,Default,,0000,0000,0000,,She randomly chooses one\Nstudent to collect homework.
Dialogue: 0,0:57:58.15,0:58:00.88,Default,,0000,0000,0000,,What's the probability of\Nselecting a female, given
Dialogue: 0,0:58:00.88,0:58:02.80,Default,,0000,0000,0000,,that she chooses a random--
Dialogue: 0,0:58:02.80,0:58:07.28,Default,,0000,0000,0000,,that she chooses randomly\Nfrom only the sophomores?
Dialogue: 0,0:58:07.28,0:58:09.98,Default,,0000,0000,0000,,So the given that part\Nis that she's only
Dialogue: 0,0:58:09.98,0:58:11.69,Default,,0000,0000,0000,,choosing from the sophomores.
Dialogue: 0,0:58:11.69,0:58:15.05,Default,,0000,0000,0000,,That means you can eliminate all\Nthe freshmen, all the juniors,
Dialogue: 0,0:58:15.05,0:58:16.03,Default,,0000,0000,0000,,all the seniors.
Dialogue: 0,0:58:16.03,0:58:21.92,Default,,0000,0000,0000,,So your sample space has reduced\Ndown to just the 21 sophomores.
Dialogue: 0,0:58:21.92,0:58:24.78,Default,,0000,0000,0000,,So now what's the probability\Nof selecting a female?
Dialogue: 0,0:58:24.78,0:58:28.79,Default,,0000,0000,0000,,We'll look within the remaining\Nchoices, those 20 mining
Dialogue: 0,0:58:28.79,0:58:29.60,Default,,0000,0000,0000,,choices.
Dialogue: 0,0:58:29.60,0:58:33.38,Default,,0000,0000,0000,,5 of them are female out\Nof the 21 sophomores.
Dialogue: 0,0:58:33.38,0:58:34.70,Default,,0000,0000,0000,,Notice again, I said it.
Dialogue: 0,0:58:34.70,0:58:36.56,Default,,0000,0000,0000,,But I'll actually write it.
Dialogue: 0,0:58:36.56,0:58:38.06,Default,,0000,0000,0000,,There are 21 sophomores.
Dialogue: 0,0:58:38.06,0:58:40.88,Default,,0000,0000,0000,,So the probability\Nthat the person--
Dialogue: 0,0:58:40.88,0:58:43.37,Default,,0000,0000,0000,,the student was female,\Ngiven that the person was
Dialogue: 0,0:58:43.37,0:58:47.45,Default,,0000,0000,0000,,a sophomore is going to\Nbe the number of females
Dialogue: 0,0:58:47.45,0:58:52.01,Default,,0000,0000,0000,,over the remaining things in the\Nsample space, which are just 21
Dialogue: 0,0:58:52.01,0:58:52.72,Default,,0000,0000,0000,,sophomores.
Dialogue: 0,0:58:52.72,0:58:56.27,Default,,0000,0000,0000,,So your answer is 5/21.
Dialogue: 0,0:58:56.27,0:59:00.47,Default,,0000,0000,0000,,Very simple if you understand\Nwhat you're trying to do.
Dialogue: 0,0:59:00.47,0:59:02.01,Default,,0000,0000,0000,,This might be a good\Nplace to mention
Dialogue: 0,0:59:02.01,0:59:05.37,Default,,0000,0000,0000,,that we can use the conditional\Nprobability ideas to modify
Dialogue: 0,0:59:05.37,0:59:07.26,Default,,0000,0000,0000,,the product rule for\Nindependent events
Dialogue: 0,0:59:07.26,0:59:11.73,Default,,0000,0000,0000,,that we've talked about\Nearlier to handle cases
Dialogue: 0,0:59:11.73,0:59:15.60,Default,,0000,0000,0000,,where the events are dependent,\Nas we've been talking about.
Dialogue: 0,0:59:15.60,0:59:18.12,Default,,0000,0000,0000,,Recall that the product\Nrule for independent events
Dialogue: 0,0:59:18.12,0:59:20.75,Default,,0000,0000,0000,,says if two events E\Nand F are independent,
Dialogue: 0,0:59:20.75,0:59:23.46,Default,,0000,0000,0000,,that the probability\Nof E and F is
Dialogue: 0,0:59:23.46,0:59:27.68,Default,,0000,0000,0000,,equal to the probability of\NE times the probability of F.
Dialogue: 0,0:59:27.68,0:59:31.29,Default,,0000,0000,0000,,But if the two\Nevents are dependent,
Dialogue: 0,0:59:31.29,0:59:35.03,Default,,0000,0000,0000,,we can modify that slightly to\Nsay that the probability of E
Dialogue: 0,0:59:35.03,0:59:38.78,Default,,0000,0000,0000,,and F is equal to the\Nprobability of E times
Dialogue: 0,0:59:38.78,0:59:42.71,Default,,0000,0000,0000,,the probability of F given\NE. It's still a product,
Dialogue: 0,0:59:42.71,0:59:44.96,Default,,0000,0000,0000,,but you're using that\Nconditional probability.
Dialogue: 0,0:59:44.96,0:59:48.52,Default,,0000,0000,0000,,
Dialogue: 0,0:59:48.52,0:59:51.17,Default,,0000,0000,0000,,The two events E\Nand F are dependent.
Dialogue: 0,0:59:51.17,0:59:54.33,Default,,0000,0000,0000,,Then the probability E and\NF is the probability of E
Dialogue: 0,0:59:54.33,1:00:01.26,Default,,0000,0000,0000,,times the probability of F,\Ngiven E. Let's do an example.
Dialogue: 0,1:00:01.26,1:00:04.01,Default,,0000,0000,0000,,Let's draw two cards from\Na standard deck of 52
Dialogue: 0,1:00:04.01,1:00:06.00,Default,,0000,0000,0000,,without replacement.
Dialogue: 0,1:00:06.00,1:00:08.81,Default,,0000,0000,0000,,What's the probability that\Nyour first card is a spade
Dialogue: 0,1:00:08.81,1:00:11.58,Default,,0000,0000,0000,,and your second car is red?
Dialogue: 0,1:00:11.58,1:00:13.38,Default,,0000,0000,0000,,Recall your deck of cards.
Dialogue: 0,1:00:13.38,1:00:15.25,Default,,0000,0000,0000,,You have to know enough\Nabout a deck of cards
Dialogue: 0,1:00:15.25,1:00:17.45,Default,,0000,0000,0000,,to answer these\Nsort of questions.
Dialogue: 0,1:00:17.45,1:00:20.16,Default,,0000,0000,0000,,The first card\Nbeing a spade causes
Dialogue: 0,1:00:20.16,1:00:22.45,Default,,0000,0000,0000,,you to look at the deck of\Ncards and realize that there
Dialogue: 0,1:00:22.45,1:00:25.23,Default,,0000,0000,0000,,are 13 spades in the deck.
Dialogue: 0,1:00:25.23,1:00:27.93,Default,,0000,0000,0000,,And the second card\Nbeing red causes
Dialogue: 0,1:00:27.93,1:00:31.50,Default,,0000,0000,0000,,you to look at the deck of cards\Nand realize that 26 of those 52
Dialogue: 0,1:00:31.50,1:00:34.08,Default,,0000,0000,0000,,cards are red.
Dialogue: 0,1:00:34.08,1:00:35.58,Default,,0000,0000,0000,,The product rule\Nfor dependent event
Dialogue: 0,1:00:35.58,1:00:38.16,Default,,0000,0000,0000,,says if two event E\Nand F are dependent,
Dialogue: 0,1:00:38.16,1:00:41.48,Default,,0000,0000,0000,,then the probability of E and\NF is the probability of E times
Dialogue: 0,1:00:41.48,1:00:44.60,Default,,0000,0000,0000,,the probability of\NF, given E. Here,
Dialogue: 0,1:00:44.60,1:00:46.92,Default,,0000,0000,0000,,that would be the probability\Nof the first being
Dialogue: 0,1:00:46.92,1:00:51.63,Default,,0000,0000,0000,,a spade and the second being a\Nred is equal to the probability
Dialogue: 0,1:00:51.63,1:00:53.61,Default,,0000,0000,0000,,that the first is a spade\Ntimes of probability
Dialogue: 0,1:00:53.61,1:00:58.87,Default,,0000,0000,0000,,that the second is red, given\Nthat the first is a spade.
Dialogue: 0,1:00:58.87,1:01:01.72,Default,,0000,0000,0000,,And that should be easy\Nenough to figure out.
Dialogue: 0,1:01:01.72,1:01:04.12,Default,,0000,0000,0000,,What is the probability\Nthat the first is a spade?
Dialogue: 0,1:01:04.12,1:01:06.88,Default,,0000,0000,0000,,Well, as we said, there\Nare 13 spades in the deck.
Dialogue: 0,1:01:06.88,1:01:09.46,Default,,0000,0000,0000,,So the probability of drawing\Na spade on that first card
Dialogue: 0,1:01:09.46,1:01:12.23,Default,,0000,0000,0000,,is 13/52.
Dialogue: 0,1:01:12.23,1:01:15.74,Default,,0000,0000,0000,,Also, keep in mind, you\Ncould reduce 13/52 to 1/4
Dialogue: 0,1:01:15.74,1:01:16.49,Default,,0000,0000,0000,,if you want to.
Dialogue: 0,1:01:16.49,1:01:18.53,Default,,0000,0000,0000,,You don't necessarily\Nhave to, but you can.
Dialogue: 0,1:01:18.53,1:01:20.36,Default,,0000,0000,0000,,What about the probability\Nof the second one
Dialogue: 0,1:01:20.36,1:01:22.64,Default,,0000,0000,0000,,being red, given that\Nthe first was a spade?
Dialogue: 0,1:01:22.64,1:01:25.01,Default,,0000,0000,0000,,This is a little tricky if\Nyou don't think because you're
Dialogue: 0,1:01:25.01,1:01:27.29,Default,,0000,0000,0000,,drawing without\Nreplacement, which
Dialogue: 0,1:01:27.29,1:01:30.80,Default,,0000,0000,0000,,means if you draw out a spade,\Nyou're not putting it back in.
Dialogue: 0,1:01:30.80,1:01:35.72,Default,,0000,0000,0000,,So the deck now only has 12\Nspades and one less card.
Dialogue: 0,1:01:35.72,1:01:38.36,Default,,0000,0000,0000,,So that would be instead\Nof 52 cards in the deck,
Dialogue: 0,1:01:38.36,1:01:40.34,Default,,0000,0000,0000,,there would only 51 in a deck.
Dialogue: 0,1:01:40.34,1:01:42.56,Default,,0000,0000,0000,,There are still 26 red cards.
Dialogue: 0,1:01:42.56,1:01:45.07,Default,,0000,0000,0000,,But now one of the\Ncards has been taken out
Dialogue: 0,1:01:45.07,1:01:45.78,Default,,0000,0000,0000,,and not replaced.
Dialogue: 0,1:01:45.78,1:01:48.29,Default,,0000,0000,0000,,So there are only 51\Ncards left in the deck.
Dialogue: 0,1:01:48.29,1:01:51.44,Default,,0000,0000,0000,,So the probability of\Ngetting a second red, given
Dialogue: 0,1:01:51.44,1:01:53.20,Default,,0000,0000,0000,,that you've taken a spade--
Dialogue: 0,1:01:53.20,1:01:58.54,Default,,0000,0000,0000,,it's not 26 out of 52, but\Ninstead is 26 out of 51.
Dialogue: 0,1:01:58.54,1:02:01.82,Default,,0000,0000,0000,,And if you multiply 1\Ntimes 26 and 4 times 51,
Dialogue: 0,1:02:01.82,1:02:04.55,Default,,0000,0000,0000,,you'll get 26 over 204.
Dialogue: 0,1:02:04.55,1:02:08.06,Default,,0000,0000,0000,,And as I said, you could\Nhave left that 1/4 are 13/52.
Dialogue: 0,1:02:08.06,1:02:09.47,Default,,0000,0000,0000,,You did not have to simplify it.
Dialogue: 0,1:02:09.47,1:02:11.81,Default,,0000,0000,0000,,But it's nicer if you notice\Nthose sorts of things,
Dialogue: 0,1:02:11.81,1:02:12.92,Default,,0000,0000,0000,,I believe.
Dialogue: 0,1:02:12.92,1:02:14.69,Default,,0000,0000,0000,,So the answer would\Nbe 13 over 102,
Dialogue: 0,1:02:14.69,1:02:16.64,Default,,0000,0000,0000,,if you simplified\Nas much as you can.
Dialogue: 0,1:02:16.64,1:02:19.40,Default,,0000,0000,0000,,Web Assign doesn't really care\Nif you simplify fractions.
Dialogue: 0,1:02:19.40,1:02:23.39,Default,,0000,0000,0000,,But if you want to, you can.
Dialogue: 0,1:02:23.39,1:02:24.59,Default,,0000,0000,0000,,How about this one?
Dialogue: 0,1:02:24.59,1:02:27.14,Default,,0000,0000,0000,,Three cards are down from a\Nwell-shuffled standard deck
Dialogue: 0,1:02:27.14,1:02:28.43,Default,,0000,0000,0000,,of 52.
Dialogue: 0,1:02:28.43,1:02:30.98,Default,,0000,0000,0000,,What is the probability\Nthat the first card is red,
Dialogue: 0,1:02:30.98,1:02:34.25,Default,,0000,0000,0000,,the second card is black,\Nand the third is red?
Dialogue: 0,1:02:34.25,1:02:36.93,Default,,0000,0000,0000,,And give your answer\Nas a fraction.
Dialogue: 0,1:02:36.93,1:02:39.84,Default,,0000,0000,0000,,Remember, 52 cards in the deck.
Dialogue: 0,1:02:39.84,1:02:48.44,Default,,0000,0000,0000,,The deck of cards has heart,\Ndiamonds, spades, and clubs.
Dialogue: 0,1:02:48.44,1:02:51.95,Default,,0000,0000,0000,,There are 2's, 3's, 4's, 5's,\N6's, 7's, 8's, 9's, and 10's.
Dialogue: 0,1:02:51.95,1:02:54.80,Default,,0000,0000,0000,,And then there's jacks, queens,\Nkings, which are the face card.
Dialogue: 0,1:02:54.80,1:02:55.73,Default,,0000,0000,0000,,And then there aces--
Dialogue: 0,1:02:55.73,1:02:58.69,Default,,0000,0000,0000,,52 cards.
Dialogue: 0,1:02:58.69,1:03:02.78,Default,,0000,0000,0000,,The product rule says that the\Nprobability that the first was
Dialogue: 0,1:03:02.78,1:03:05.93,Default,,0000,0000,0000,,red, and the second one's\Nblack, and third one's red,
Dialogue: 0,1:03:05.93,1:03:08.74,Default,,0000,0000,0000,,is the probability the first\Nwas red times the probability
Dialogue: 0,1:03:08.74,1:03:11.12,Default,,0000,0000,0000,,that the second one's black,\Ngiven that the first one was
Dialogue: 0,1:03:11.12,1:03:14.09,Default,,0000,0000,0000,,red, and times the probability\Nthe third one is red,
Dialogue: 0,1:03:14.09,1:03:17.14,Default,,0000,0000,0000,,given that the first\Ntwo choices were made.
Dialogue: 0,1:03:17.14,1:03:19.39,Default,,0000,0000,0000,,So those last two are\Nconditional probabilities.
Dialogue: 0,1:03:19.39,1:03:21.10,Default,,0000,0000,0000,,It's basically the\Nprobability of getting
Dialogue: 0,1:03:21.10,1:03:22.48,Default,,0000,0000,0000,,a red, the\Nprobability of getting
Dialogue: 0,1:03:22.48,1:03:24.73,Default,,0000,0000,0000,,a black, the probability\Nof getting another red.
Dialogue: 0,1:03:24.73,1:03:27.11,Default,,0000,0000,0000,,But you're taking\Ninto consideration
Dialogue: 0,1:03:27.11,1:03:28.36,Default,,0000,0000,0000,,what's already been drawn out.
Dialogue: 0,1:03:28.36,1:03:32.09,Default,,0000,0000,0000,,Because if you're dealing cards,\Nthey're not being put back in.
Dialogue: 0,1:03:32.09,1:03:33.34,Default,,0000,0000,0000,,So they're not going back in.
Dialogue: 0,1:03:33.34,1:03:35.68,Default,,0000,0000,0000,,So that means it's\Nwithout replacement.
Dialogue: 0,1:03:35.68,1:03:38.26,Default,,0000,0000,0000,,That's what makes the\Nconditional probability.
Dialogue: 0,1:03:38.26,1:03:41.26,Default,,0000,0000,0000,,So anyway, we want to find\Nout what is the probability
Dialogue: 0,1:03:41.26,1:03:42.60,Default,,0000,0000,0000,,of going that first card red?
Dialogue: 0,1:03:42.60,1:03:46.29,Default,,0000,0000,0000,,Well, there are 26\Ncards in the deck of 52.
Dialogue: 0,1:03:46.29,1:03:48.16,Default,,0000,0000,0000,,So that's 26 over 52.
Dialogue: 0,1:03:48.16,1:03:49.74,Default,,0000,0000,0000,,That also happens to be 1/2.
Dialogue: 0,1:03:49.74,1:03:52.84,Default,,0000,0000,0000,,So if you wanted to write it\Nis 1/2, that's perfectly OK.
Dialogue: 0,1:03:52.84,1:03:54.91,Default,,0000,0000,0000,,I'll leave it that way\Nfor the time being.
Dialogue: 0,1:03:54.91,1:03:56.98,Default,,0000,0000,0000,,But what about drawing\Na second black?
Dialogue: 0,1:03:56.98,1:03:59.05,Default,,0000,0000,0000,,Now you've got to be careful.
Dialogue: 0,1:03:59.05,1:04:02.53,Default,,0000,0000,0000,,After that first card is\Ndealt, it doesn't go back in.
Dialogue: 0,1:04:02.53,1:04:07.00,Default,,0000,0000,0000,,So there are only 51\Ncards left in the deck.
Dialogue: 0,1:04:07.00,1:04:08.10,Default,,0000,0000,0000,,I drew out a red.
Dialogue: 0,1:04:08.10,1:04:10.70,Default,,0000,0000,0000,,So there are only 25\Nreds and 26 blacks.
Dialogue: 0,1:04:10.70,1:04:13.73,Default,,0000,0000,0000,,So the probability of getting\Na second black from that deck
Dialogue: 0,1:04:13.73,1:04:17.06,Default,,0000,0000,0000,,of 51 cards is 26 out of 51.
Dialogue: 0,1:04:17.06,1:04:17.87,Default,,0000,0000,0000,,I wanted black.
Dialogue: 0,1:04:17.87,1:04:19.67,Default,,0000,0000,0000,,And there are still\N26 blacks in there.
Dialogue: 0,1:04:19.67,1:04:22.06,Default,,0000,0000,0000,,But there are only\N51 cards in the deck.
Dialogue: 0,1:04:22.06,1:04:23.93,Default,,0000,0000,0000,,So the probability of\Ngetting a second black,
Dialogue: 0,1:04:23.93,1:04:28.34,Default,,0000,0000,0000,,given that the first one\Nred is red, is 26 out of 51.
Dialogue: 0,1:04:28.34,1:04:30.32,Default,,0000,0000,0000,,And finally, what\Nis the probability
Dialogue: 0,1:04:30.32,1:04:33.60,Default,,0000,0000,0000,,the third one is red, given\Nthat the other two were chosen?
Dialogue: 0,1:04:33.60,1:04:35.52,Default,,0000,0000,0000,,Well, now you've taken\Ntwo cards out the deck.
Dialogue: 0,1:04:35.52,1:04:37.25,Default,,0000,0000,0000,,So there are only 50 cards.
Dialogue: 0,1:04:37.25,1:04:40.18,Default,,0000,0000,0000,,And you've taken one of each\Ncolor, one red, one black.
Dialogue: 0,1:04:40.18,1:04:43.55,Default,,0000,0000,0000,,So there only 25 reds and\N25 blacks left in the deck.
Dialogue: 0,1:04:43.55,1:04:50.17,Default,,0000,0000,0000,,So the probability that the\Nnext one is red is 25 out of 50.
Dialogue: 0,1:04:50.17,1:04:52.58,Default,,0000,0000,0000,,And according to the product\Nrule, all you gotta do
Dialogue: 0,1:04:52.58,1:04:53.70,Default,,0000,0000,0000,,is multiply those together.
Dialogue: 0,1:04:53.70,1:04:57.90,Default,,0000,0000,0000,,
Dialogue: 0,1:04:57.90,1:04:59.77,Default,,0000,0000,0000,,Given your answer is a\Nfraction-- if you just
Dialogue: 0,1:04:59.77,1:05:06.13,Default,,0000,0000,0000,,multiply straight across, you'll\Nget 16,900 divided by 132,600.
Dialogue: 0,1:05:06.13,1:05:08.20,Default,,0000,0000,0000,,Now you can simplify\Nthat any way you want to.
Dialogue: 0,1:05:08.20,1:05:09.70,Default,,0000,0000,0000,,In fact, you could\Nhave gone back up
Dialogue: 0,1:05:09.70,1:05:13.21,Default,,0000,0000,0000,,to the top and simplified 26\Nout of 52 as being a half,
Dialogue: 0,1:05:13.21,1:05:15.34,Default,,0000,0000,0000,,and 25 out of 50 being a half.
Dialogue: 0,1:05:15.34,1:05:17.38,Default,,0000,0000,0000,,And you could have gotten\Na lot nicer answer.
Dialogue: 0,1:05:17.38,1:05:19.78,Default,,0000,0000,0000,,But as said, Web Assign\Nreally doesn't care.
Dialogue: 0,1:05:19.78,1:05:21.76,Default,,0000,0000,0000,,But if you did want\Nto simplify, you
Dialogue: 0,1:05:21.76,1:05:28.25,Default,,0000,0000,0000,,could get it all the\Nway down to 13 over 102.
Dialogue: 0,1:05:28.25,1:05:31.82,Default,,0000,0000,0000,,I want to wrap up this section\Nby going back to three problems
Dialogue: 0,1:05:31.82,1:05:34.50,Default,,0000,0000,0000,,that I covered in section 12.4.
Dialogue: 0,1:05:34.50,1:05:37.01,Default,,0000,0000,0000,,And at the time, I said\Nthat I would show you
Dialogue: 0,1:05:37.01,1:05:38.72,Default,,0000,0000,0000,,what I consider to\Nbe an easier way
Dialogue: 0,1:05:38.72,1:05:40.77,Default,,0000,0000,0000,,to solve those three problems.
Dialogue: 0,1:05:40.77,1:05:42.16,Default,,0000,0000,0000,,So let's do that.
Dialogue: 0,1:05:42.16,1:05:44.54,Default,,0000,0000,0000,,If you look in the upper right\Nhand corner of the screen,
Dialogue: 0,1:05:44.54,1:05:48.39,Default,,0000,0000,0000,,you'll see I brought up a screen\Ncapture of that problem that
Dialogue: 0,1:05:48.39,1:05:51.62,Default,,0000,0000,0000,,I'm talking about-- at least\Nthe first one of those--
Dialogue: 0,1:05:51.62,1:05:54.14,Default,,0000,0000,0000,,and how we solved it before.
Dialogue: 0,1:05:54.14,1:05:57.08,Default,,0000,0000,0000,,That problem said if\Nwe roll a standard die
Dialogue: 0,1:05:57.08,1:05:59.30,Default,,0000,0000,0000,,three times, what is\Nthe probability that we
Dialogue: 0,1:05:59.30,1:06:02.31,Default,,0000,0000,0000,,get at least one 4.
Dialogue: 0,1:06:02.31,1:06:05.61,Default,,0000,0000,0000,,Instead of solving it the\Nway I did in section 4,
Dialogue: 0,1:06:05.61,1:06:08.52,Default,,0000,0000,0000,,I want to solve it using\Nthe complement rule, which
Dialogue: 0,1:06:08.52,1:06:12.06,Default,,0000,0000,0000,,we did use in section 4,\Ncombined with the product
Dialogue: 0,1:06:12.06,1:06:14.52,Default,,0000,0000,0000,,rule for probability that\Nwe covered in this section.
Dialogue: 0,1:06:14.52,1:06:16.48,Default,,0000,0000,0000,,I think you'll find\Nthis much easier.
Dialogue: 0,1:06:16.48,1:06:20.49,Default,,0000,0000,0000,,To do that, I start off with a\Ncomplement rule just at the 4.
Dialogue: 0,1:06:20.49,1:06:26.02,Default,,0000,0000,0000,,In this case, the probability\Nof getting at least one 4 is 1
Dialogue: 0,1:06:26.02,1:06:29.47,Default,,0000,0000,0000,,minus the probability of not\Ngetting any 4's or getting zero
Dialogue: 0,1:06:29.47,1:06:30.88,Default,,0000,0000,0000,,4's.
Dialogue: 0,1:06:30.88,1:06:33.78,Default,,0000,0000,0000,,Now the probability\Nof getting zero 4's
Dialogue: 0,1:06:33.78,1:06:35.50,Default,,0000,0000,0000,,is the key to this\Nwhole calculation.
Dialogue: 0,1:06:35.50,1:06:38.35,Default,,0000,0000,0000,,Because once we get that,\Nwe just say 1 minus that,
Dialogue: 0,1:06:38.35,1:06:40.46,Default,,0000,0000,0000,,and we'll have the\Nanswer we're looking for.
Dialogue: 0,1:06:40.46,1:06:44.14,Default,,0000,0000,0000,,But the probability of\Ngetting no 4's, according
Dialogue: 0,1:06:44.14,1:06:46.12,Default,,0000,0000,0000,,to the product rule\Nof probability,
Dialogue: 0,1:06:46.12,1:06:48.64,Default,,0000,0000,0000,,is simply the probability\Nof getting no 4's
Dialogue: 0,1:06:48.64,1:06:51.49,Default,,0000,0000,0000,,on roll one times the\Nprobability of getting no 4
Dialogue: 0,1:06:51.49,1:06:54.76,Default,,0000,0000,0000,,on roll two times the\Nprobability of getting no 4
Dialogue: 0,1:06:54.76,1:06:55.87,Default,,0000,0000,0000,,on roll three.
Dialogue: 0,1:06:55.87,1:06:57.76,Default,,0000,0000,0000,,That's what we learned\Nin this section.
Dialogue: 0,1:06:57.76,1:07:01.09,Default,,0000,0000,0000,,But each of those probabilities\Nis easy to calculate,
Dialogue: 0,1:07:01.09,1:07:04.00,Default,,0000,0000,0000,,because if you don't\Nwant to get a 4,
Dialogue: 0,1:07:04.00,1:07:06.22,Default,,0000,0000,0000,,you just have to get\Nanything but a 4.
Dialogue: 0,1:07:06.22,1:07:10.55,Default,,0000,0000,0000,,And also, because you're rolling\Ndie, the rolls are independent.
Dialogue: 0,1:07:10.55,1:07:12.46,Default,,0000,0000,0000,,So the probabilities\Ndon't change.
Dialogue: 0,1:07:12.46,1:07:14.87,Default,,0000,0000,0000,,So those will be\Nthree probabilities
Dialogue: 0,1:07:14.87,1:07:17.29,Default,,0000,0000,0000,,that are exactly identical\Nmultiplied together.
Dialogue: 0,1:07:17.29,1:07:20.83,Default,,0000,0000,0000,,Well, what is the probability\Nof not getting a 4 on roll one?
Dialogue: 0,1:07:20.83,1:07:22.68,Default,,0000,0000,0000,,Think about it.
Dialogue: 0,1:07:22.68,1:07:24.47,Default,,0000,0000,0000,,If you're not going to\Nget a 4 on roll one,
Dialogue: 0,1:07:24.47,1:07:27.59,Default,,0000,0000,0000,,you can get either\Na 1, 2, 3, 5, or 6.
Dialogue: 0,1:07:27.59,1:07:30.14,Default,,0000,0000,0000,,That's five possibilities\Nout of six possibilities.
Dialogue: 0,1:07:30.14,1:07:33.02,Default,,0000,0000,0000,,So the probability of\Nthat happening is 5/6.
Dialogue: 0,1:07:33.02,1:07:35.40,Default,,0000,0000,0000,,And because the rolls\Nare independent,
Dialogue: 0,1:07:35.40,1:07:38.09,Default,,0000,0000,0000,,it's going to be 5/6\Non the second roll
Dialogue: 0,1:07:38.09,1:07:39.45,Default,,0000,0000,0000,,and on the third roll.
Dialogue: 0,1:07:39.45,1:07:41.90,Default,,0000,0000,0000,,So the probability\Nof getting no 4's
Dialogue: 0,1:07:41.90,1:07:45.98,Default,,0000,0000,0000,,on any of those three rolls,\Nis simply 5/6 times 5/6 times
Dialogue: 0,1:07:45.98,1:07:51.63,Default,,0000,0000,0000,,5/6, which is 125 over 216.
Dialogue: 0,1:07:51.63,1:07:53.34,Default,,0000,0000,0000,,Remember, we want to\Nknow the probability
Dialogue: 0,1:07:53.34,1:07:55.02,Default,,0000,0000,0000,,of getting at least one 4.
Dialogue: 0,1:07:55.02,1:07:57.99,Default,,0000,0000,0000,,So we have to take that\Nand subtract it from 1.
Dialogue: 0,1:07:57.99,1:08:02.22,Default,,0000,0000,0000,,So you have 1\Nminus 125 over 216,
Dialogue: 0,1:08:02.22,1:08:06.66,Default,,0000,0000,0000,,which leaves 91 out of\N216 as your final answer.
Dialogue: 0,1:08:06.66,1:08:09.16,Default,,0000,0000,0000,,And if you look up in the corner\Nof the screen on the right,
Dialogue: 0,1:08:09.16,1:08:11.37,Default,,0000,0000,0000,,you'll see that's exactly\Nthe answer we got before.
Dialogue: 0,1:08:11.37,1:08:14.10,Default,,0000,0000,0000,,I think this solution\Nis much easier.
Dialogue: 0,1:08:14.10,1:08:16.35,Default,,0000,0000,0000,,Let's do another one.
Dialogue: 0,1:08:16.35,1:08:18.27,Default,,0000,0000,0000,,Up in the upper\Nright hand corner,
Dialogue: 0,1:08:18.27,1:08:20.40,Default,,0000,0000,0000,,I put up the\Nsolution that we used
Dialogue: 0,1:08:20.40,1:08:22.66,Default,,0000,0000,0000,,to solve this problem in 12.4.
Dialogue: 0,1:08:22.66,1:08:25.29,Default,,0000,0000,0000,,And you can see the answer\Nin the solution technique.
Dialogue: 0,1:08:25.29,1:08:28.44,Default,,0000,0000,0000,,I'm going to do the same\Nthing with this problem.
Dialogue: 0,1:08:28.44,1:08:29.82,Default,,0000,0000,0000,,This time we're drawing cards.
Dialogue: 0,1:08:29.82,1:08:31.65,Default,,0000,0000,0000,,We're drawing three\Ncards from standard deck
Dialogue: 0,1:08:31.65,1:08:33.14,Default,,0000,0000,0000,,without replacement.
Dialogue: 0,1:08:33.14,1:08:35.43,Default,,0000,0000,0000,,And we want to know the\Nprobability of getting at least
Dialogue: 0,1:08:35.43,1:08:39.06,Default,,0000,0000,0000,,one that's a face card.
Dialogue: 0,1:08:39.06,1:08:41.65,Default,,0000,0000,0000,,We'll start off just before\Nusing the complement rule.
Dialogue: 0,1:08:41.65,1:08:44.64,Default,,0000,0000,0000,,But I want to do the rest of\Nthe problem using the product
Dialogue: 0,1:08:44.64,1:08:48.22,Default,,0000,0000,0000,,rule for probability that\Nwe learned in this section.
Dialogue: 0,1:08:48.22,1:08:50.88,Default,,0000,0000,0000,,So the complement rule\Nhere says the probability
Dialogue: 0,1:08:50.88,1:08:54.41,Default,,0000,0000,0000,,of getting at least one face\Ncard is 1 minus the probability
Dialogue: 0,1:08:54.41,1:08:57.43,Default,,0000,0000,0000,,that we don't\Ngetting face cards.
Dialogue: 0,1:08:57.43,1:08:59.10,Default,,0000,0000,0000,,But remember, we're\Ndrawing three times.
Dialogue: 0,1:08:59.10,1:09:00.68,Default,,0000,0000,0000,,And we're drawing\Nwithout replacement.
Dialogue: 0,1:09:00.68,1:09:03.69,Default,,0000,0000,0000,,So keep that in the\Nback of your mind.
Dialogue: 0,1:09:03.69,1:09:06.87,Default,,0000,0000,0000,,The probability of getting\Nzero face cards on three
Dialogue: 0,1:09:06.87,1:09:09.90,Default,,0000,0000,0000,,draws by the product\Nrule is the probability
Dialogue: 0,1:09:09.90,1:09:12.57,Default,,0000,0000,0000,,we don't get a face card in the\Nfirst draw times of probability
Dialogue: 0,1:09:12.57,1:09:14.43,Default,,0000,0000,0000,,we don't get a face\Ncard on a second draw
Dialogue: 0,1:09:14.43,1:09:16.14,Default,,0000,0000,0000,,times the probability\Nwe don't get a face
Dialogue: 0,1:09:16.14,1:09:17.89,Default,,0000,0000,0000,,card on the third draw.
Dialogue: 0,1:09:17.89,1:09:20.58,Default,,0000,0000,0000,,But because these draws\Nare without replacement,
Dialogue: 0,1:09:20.58,1:09:22.08,Default,,0000,0000,0000,,these are conditional\Nprobabilities.
Dialogue: 0,1:09:22.08,1:09:24.42,Default,,0000,0000,0000,,And they're going to change\Nbetween draws because you're
Dialogue: 0,1:09:24.42,1:09:26.32,Default,,0000,0000,0000,,not putting the card back in.
Dialogue: 0,1:09:26.32,1:09:28.06,Default,,0000,0000,0000,,So keep that in mind.
Dialogue: 0,1:09:28.06,1:09:29.97,Default,,0000,0000,0000,,What is the probability\Nof not getting a face
Dialogue: 0,1:09:29.97,1:09:31.15,Default,,0000,0000,0000,,card on the first card?
Dialogue: 0,1:09:31.15,1:09:33.40,Default,,0000,0000,0000,,Well, if you look back in\Nthe upper right hand corner,
Dialogue: 0,1:09:33.40,1:09:35.16,Default,,0000,0000,0000,,you'll see that we\Ntalked about this.
Dialogue: 0,1:09:35.16,1:09:36.69,Default,,0000,0000,0000,,There are 52 cards in the deck.
Dialogue: 0,1:09:36.69,1:09:37.85,Default,,0000,0000,0000,,There are 12 face cards.
Dialogue: 0,1:09:37.85,1:09:40.71,Default,,0000,0000,0000,,There are 40 cards that\Nare not face cards.
Dialogue: 0,1:09:40.71,1:09:43.02,Default,,0000,0000,0000,,So the probability of\Nnot getting a face card
Dialogue: 0,1:09:43.02,1:09:47.04,Default,,0000,0000,0000,,on the first draw is\Nsimply 40 out of 52.
Dialogue: 0,1:09:47.04,1:09:49.17,Default,,0000,0000,0000,,But when we go to calculate\Nthe second probability,
Dialogue: 0,1:09:49.17,1:09:51.72,Default,,0000,0000,0000,,the probability we dong get\Na face card on a second draw,
Dialogue: 0,1:09:51.72,1:09:53.43,Default,,0000,0000,0000,,this is a conditional\Nprobability
Dialogue: 0,1:09:53.43,1:09:55.86,Default,,0000,0000,0000,,because we're not\Nputting the card back in.
Dialogue: 0,1:09:55.86,1:09:58.44,Default,,0000,0000,0000,,So we got a non-face\Ncard on the first draws.
Dialogue: 0,1:09:58.44,1:10:01.82,Default,,0000,0000,0000,,So that leaves only\N39 of them out of 51,
Dialogue: 0,1:10:01.82,1:10:03.61,Default,,0000,0000,0000,,because we're not\Nputting the card back in.
Dialogue: 0,1:10:03.61,1:10:05.27,Default,,0000,0000,0000,,So the probability\Non the second draw
Dialogue: 0,1:10:05.27,1:10:06.96,Default,,0000,0000,0000,,of not getting a face card--
Dialogue: 0,1:10:06.96,1:10:10.26,Default,,0000,0000,0000,,well, there are only 39 face\Ncards left out of 51 cards.
Dialogue: 0,1:10:10.26,1:10:11.64,Default,,0000,0000,0000,,And then you can\Nsee the pattern.
Dialogue: 0,1:10:11.64,1:10:13.44,Default,,0000,0000,0000,,By the time you get to\Nthe third face card,
Dialogue: 0,1:10:13.44,1:10:15.91,Default,,0000,0000,0000,,you're down to 38\Nface cards in the deck
Dialogue: 0,1:10:15.91,1:10:17.92,Default,,0000,0000,0000,,with only 50 cards in it.
Dialogue: 0,1:10:17.92,1:10:20.70,Default,,0000,0000,0000,,So the probability of getting no\Nface cards by the product rule
Dialogue: 0,1:10:20.70,1:10:24.87,Default,,0000,0000,0000,,is simply the product of\N40/52 times 39/51 times
Dialogue: 0,1:10:24.87,1:10:32.82,Default,,0000,0000,0000,,38/50, which comes out to\Nbe 59,280 over 132,600.
Dialogue: 0,1:10:32.82,1:10:36.48,Default,,0000,0000,0000,,Now this particular problem\Nask us to calculate and leave
Dialogue: 0,1:10:36.48,1:10:38.88,Default,,0000,0000,0000,,the answer as a percent\Nrounded to one decimal place.
Dialogue: 0,1:10:38.88,1:10:41.68,Default,,0000,0000,0000,,So we're going to have to\Nchange this to a decimal.
Dialogue: 0,1:10:41.68,1:10:43.80,Default,,0000,0000,0000,,So this is probably\Na good time to do it.
Dialogue: 0,1:10:43.80,1:10:46.35,Default,,0000,0000,0000,,You really have the probability\Nof getting at least one face
Dialogue: 0,1:10:46.35,1:10:48.65,Default,,0000,0000,0000,,card being 1 minus\Nthe probability
Dialogue: 0,1:10:48.65,1:10:50.07,Default,,0000,0000,0000,,that you don't get\Nany face cards.
Dialogue: 0,1:10:50.07,1:10:55.35,Default,,0000,0000,0000,,And we just calculate that\Nto be 59,280 over 132,600.
Dialogue: 0,1:10:55.35,1:10:58.36,Default,,0000,0000,0000,,Probably a good time to change\Nthat fraction to a decimal.
Dialogue: 0,1:10:58.36,1:11:00.60,Default,,0000,0000,0000,,We're going to change\Nthe answer to a percent,
Dialogue: 0,1:11:00.60,1:11:02.34,Default,,0000,0000,0000,,and then we're\Ngoing to round it.
Dialogue: 0,1:11:02.34,1:11:06.52,Default,,0000,0000,0000,,So carry plenty\Nof decimal places.
Dialogue: 0,1:11:06.52,1:11:08.56,Default,,0000,0000,0000,,And then when you\Nround later, you
Dialogue: 0,1:11:08.56,1:11:11.20,Default,,0000,0000,0000,,will be accurate to one\Ndecimal place as a percent.
Dialogue: 0,1:11:11.20,1:11:13.66,Default,,0000,0000,0000,,So carry plenty\Nof decimal places.
Dialogue: 0,1:11:13.66,1:11:16.21,Default,,0000,0000,0000,,And I carried it to five.
Dialogue: 0,1:11:16.21,1:11:20.23,Default,,0000,0000,0000,,So when I divided\N59,280 about 132,600,
Dialogue: 0,1:11:20.23,1:11:24.99,Default,,0000,0000,0000,,I got 0.44706 to\Nfive decimal places.
Dialogue: 0,1:11:24.99,1:11:29.25,Default,,0000,0000,0000,,1 minus that is about 0.55294.
Dialogue: 0,1:11:29.25,1:11:33.49,Default,,0000,0000,0000,,But remember, the problem said\Nto change it to a percent,
Dialogue: 0,1:11:33.49,1:11:35.26,Default,,0000,0000,0000,,and then round it to\None decimal place.
Dialogue: 0,1:11:35.26,1:11:37.96,Default,,0000,0000,0000,,Changing to a percent involves\Nmoving the decimal place
Dialogue: 0,1:11:37.96,1:11:39.46,Default,,0000,0000,0000,,two places to the right.
Dialogue: 0,1:11:39.46,1:11:43.18,Default,,0000,0000,0000,,So that become 55.294%.
Dialogue: 0,1:11:43.18,1:11:45.58,Default,,0000,0000,0000,,And now we want to round\Nit to one decimal place.
Dialogue: 0,1:11:45.58,1:11:48.04,Default,,0000,0000,0000,,So the 0.2 becomes 0.3.
Dialogue: 0,1:11:48.04,1:11:50.33,Default,,0000,0000,0000,,And if you'll look up into\Nthe upper right hand corner,
Dialogue: 0,1:11:50.33,1:11:52.96,Default,,0000,0000,0000,,you'll see it's exactly\Nthe answer we got before.
Dialogue: 0,1:11:52.96,1:11:57.36,Default,,0000,0000,0000,,And I would contend that\Nthis solution is much easier.
Dialogue: 0,1:11:57.36,1:11:59.21,Default,,0000,0000,0000,,Let's do one more.
Dialogue: 0,1:11:59.21,1:12:02.03,Default,,0000,0000,0000,,From the earlier\Nsection, we did a problem
Dialogue: 0,1:12:02.03,1:12:05.40,Default,,0000,0000,0000,,about rolling a pair\Ndice three times,
Dialogue: 0,1:12:05.40,1:12:07.95,Default,,0000,0000,0000,,and finding the probability\Nthat we get a sum of 6
Dialogue: 0,1:12:07.95,1:12:08.90,Default,,0000,0000,0000,,at least once.
Dialogue: 0,1:12:08.90,1:12:10.65,Default,,0000,0000,0000,,And again, we're going\Nto round our answer
Dialogue: 0,1:12:10.65,1:12:14.12,Default,,0000,0000,0000,,to three decimal places here.
Dialogue: 0,1:12:14.12,1:12:16.23,Default,,0000,0000,0000,,We'll start off with\Nthe complement rule,
Dialogue: 0,1:12:16.23,1:12:18.29,Default,,0000,0000,0000,,just as before back in\Nthe previous section.
Dialogue: 0,1:12:18.29,1:12:21.08,Default,,0000,0000,0000,,But again, I want to use the\Nproduct rule for probability
Dialogue: 0,1:12:21.08,1:12:25.46,Default,,0000,0000,0000,,to solve this problem instead\Nof the section 4 technique.
Dialogue: 0,1:12:25.46,1:12:27.62,Default,,0000,0000,0000,,So according to complement\Nrule, the probability
Dialogue: 0,1:12:27.62,1:12:29.90,Default,,0000,0000,0000,,of getting at least\None sum of 6 is
Dialogue: 0,1:12:29.90,1:12:32.80,Default,,0000,0000,0000,,1 minus the probability\Nof getting no sums of 6.
Dialogue: 0,1:12:32.80,1:12:35.40,Default,,0000,0000,0000,,
Dialogue: 0,1:12:35.40,1:12:37.86,Default,,0000,0000,0000,,So the key to this whole problem\Nis finding the probability
Dialogue: 0,1:12:37.86,1:12:40.11,Default,,0000,0000,0000,,of getting no sums of 6.
Dialogue: 0,1:12:40.11,1:12:41.76,Default,,0000,0000,0000,,According to the\Nproduct rule, that
Dialogue: 0,1:12:41.76,1:12:44.37,Default,,0000,0000,0000,,will just be the probability\Nof not getting a sum of 6
Dialogue: 0,1:12:44.37,1:12:48.30,Default,,0000,0000,0000,,in the first roll, not getting\Na sum of 6 on the second roll,
Dialogue: 0,1:12:48.30,1:12:50.46,Default,,0000,0000,0000,,and not getting a sum\Nof 6 on the third roll.
Dialogue: 0,1:12:50.46,1:12:53.46,Default,,0000,0000,0000,,Remember, we're\Nrolling two dice.
Dialogue: 0,1:12:53.46,1:12:56.55,Default,,0000,0000,0000,,And you've got to remember\Nwhen you're rolling two dice,
Dialogue: 0,1:12:56.55,1:12:58.41,Default,,0000,0000,0000,,you go back to\Nyour sample space.
Dialogue: 0,1:12:58.41,1:12:59.91,Default,,0000,0000,0000,,There are 36 possibilities.
Dialogue: 0,1:12:59.91,1:13:02.31,Default,,0000,0000,0000,,And we've been putting\Nthat in a table.
Dialogue: 0,1:13:02.31,1:13:05.61,Default,,0000,0000,0000,,So 36 possibilities-- and if\Nyou don't want a sum of 6,
Dialogue: 0,1:13:05.61,1:13:09.48,Default,,0000,0000,0000,,you can see where I've circled\Nall the sums that are not 6.
Dialogue: 0,1:13:09.48,1:13:11.52,Default,,0000,0000,0000,,And the only ones\Nthat aren't circled
Dialogue: 0,1:13:11.52,1:13:16.68,Default,,0000,0000,0000,,are 5, 1; 4, 2; 3,\N3; 2, 4; and 1, 5.
Dialogue: 0,1:13:16.68,1:13:17.55,Default,,0000,0000,0000,,That's five.
Dialogue: 0,1:13:17.55,1:13:18.99,Default,,0000,0000,0000,,But there are 36 together.
Dialogue: 0,1:13:18.99,1:13:21.57,Default,,0000,0000,0000,,That means there\Nare 31 that are not
Dialogue: 0,1:13:21.57,1:13:26.72,Default,,0000,0000,0000,,sums of 6's over a\Npossible 36 sums,
Dialogue: 0,1:13:26.72,1:13:28.77,Default,,0000,0000,0000,,irregardless of whether\Nit's a 6 or not.
Dialogue: 0,1:13:28.77,1:13:33.55,Default,,0000,0000,0000,,So the probability of getting\Nno sum of 6 is 31 over 36.
Dialogue: 0,1:13:33.55,1:13:36.64,Default,,0000,0000,0000,,And as in the\Nproblem before last,
Dialogue: 0,1:13:36.64,1:13:39.04,Default,,0000,0000,0000,,these rolls are independent.
Dialogue: 0,1:13:39.04,1:13:42.31,Default,,0000,0000,0000,,Rolling something on a\Nfirst roll of those two dice
Dialogue: 0,1:13:42.31,1:13:45.20,Default,,0000,0000,0000,,has no effect on the sum\Nyou get when you roll again.
Dialogue: 0,1:13:45.20,1:13:46.41,Default,,0000,0000,0000,,Those are independent events.
Dialogue: 0,1:13:46.41,1:13:47.83,Default,,0000,0000,0000,,So the probabilities\Ndon't change.
Dialogue: 0,1:13:47.83,1:13:53.34,Default,,0000,0000,0000,,So it's going to be 31 over\N36 times another 31 over 36
Dialogue: 0,1:13:53.34,1:13:55.71,Default,,0000,0000,0000,,times a third 31 over 36.
Dialogue: 0,1:13:55.71,1:14:03.51,Default,,0000,0000,0000,,And if you multiply that out,\Nyou'll get 29,791 over 46,656.
Dialogue: 0,1:14:03.51,1:14:06.48,Default,,0000,0000,0000,,So the probability of\Ngetting at least one sum of 6
Dialogue: 0,1:14:06.48,1:14:10.20,Default,,0000,0000,0000,,is 1 minus that fraction.
Dialogue: 0,1:14:10.20,1:14:13.12,Default,,0000,0000,0000,,Again, they're asking\Nus to leave the answer
Dialogue: 0,1:14:13.12,1:14:15.70,Default,,0000,0000,0000,,as a decimal rounded to\Nthree decimal places.
Dialogue: 0,1:14:15.70,1:14:18.99,Default,,0000,0000,0000,,So we want to change\Nthat to a decimal.
Dialogue: 0,1:14:18.99,1:14:20.75,Default,,0000,0000,0000,,And carry plenty\Nof decimal places
Dialogue: 0,1:14:20.75,1:14:22.17,Default,,0000,0000,0000,,so you don't get\Na rounding error.
Dialogue: 0,1:14:22.17,1:14:28.59,Default,,0000,0000,0000,,That comes out to be\Nabout 1 minus 0.63852.
Dialogue: 0,1:14:28.59,1:14:32.94,Default,,0000,0000,0000,,And that's about 0.36148.
Dialogue: 0,1:14:32.94,1:14:34.59,Default,,0000,0000,0000,,And three decimal places--
Dialogue: 0,1:14:34.59,1:14:38.76,Default,,0000,0000,0000,,that would be 0.361.
Dialogue: 0,1:14:38.76,1:14:41.01,Default,,0000,0000,0000,,If you look back in the\Nupper right hand corner,
Dialogue: 0,1:14:41.01,1:14:44.18,Default,,0000,0000,0000,,that's exactly what\Nwe got the other way.
Dialogue: 0,1:14:44.18,1:14:47.64,Default,,0000,0000,0000,,And again, my contention\Nis the 12.5 technique
Dialogue: 0,1:14:47.64,1:14:54.50,Default,,0000,0000,0000,,for these problems is much\Neasier than the 12.4 technique.
Dialogue: 0,1:14:54.50,1:14:55.00,Default,,0000,0000,0000,,