Today we'll be covering section 12.4 on the addition
and complement rules.
Before we begin, I want to say that most
of the problems in this section can be solved using techniques
we already know.
But occasionally, we'll come across certain problems
when we need to learn a little bit more about probability.
And occasionally, we'll find out that it's
convenient to use some more formulas
and have them in our tool box.
So I'm going to do something that I don't usually do.
I'm going to go ahead and give you a listing of the formulas
that we're going to add to our tool box
in advance of when we've actually talked about them.
I think it would be good to let you
look at what we're going to be using, mainly
for the reason that said earlier.
We don't always even need them.
But they will be here we do need them.
The first is called the addition or union
rule for probabilities.
It simply says that the probability of E or F
is equal to the probability of E plus the probability of F
minus the probability of E and F. And remember, or means union
and and means intersection.
The complement rule-- if E is an event,
and E with a superscript c is the complement of the event,
then the probability of E compliment is equal to 1 minus
the probability of E. The product rule for independent
and dependent events--
if two events E and F are independent,
then the probability of E and F is simply the probability of E
times the probability of F. Now the dependent version
of this rule uses a notation we haven't even introduced yet,
but I'm going to go ahead and do it anyway.
We'll come back to it later and talk more
about what the new notation means.
But it says, if two events in F are dependent,
then the probability of E and F is
equal to the probability of E times
the probability of F given E, which is also
equal to the probability of F times
the probability of E given F. Remember,
I haven't even introduced that notation
whether the vertical line is used here in this context yet,
but I will.
That leads to something called conditional probability.
And I've got two formulas related to that.
One says the probability of E given F
is equal to the probability of E and F
divided by the probability of F. Remember when I say E and F,
and means intersection.
Also, the probability of F given E
is the probability of E and F divided
by the probability of E.
Before I leave this, I told you I'm going
to rotate back to this later.
But I will say if you use that notation
with the vertical line, it's read
the probability of E given F. The vertical line is
read as given or given that.
So of course, the probability of F given E when the E and the F
are flipped around the vertical line.
Like I said, we'll come back to this later.
But this is the addition to our tool box of formulas.
Sometimes we'll need them.
Sometimes we'll just be able to go back and do
these problems using the things we already had.
But they're there if you need them.
Suppose we want to calculate the probability of rolling two dice
and getting a sum of either 5 or 9.
And you notice, I call this getting
a sum of 5 an event called E sub 1,
and the sum of getting a 9 an event called E sub 2.
That's just for convenience.
We're trying to roll two dice and find
the probability of getting a sum of either 5 or 9.
You really don't need any of the new formulas to do this.
This is just like a problem we might
have been able to work earlier.
And I want to emphasize that.
What you really need to do is build your 2 by 2 table
for rolling two dice, which gives you the sample space.
And then look.
We're looking for a sum of 5.
That's the event I call E sub 1.
There are four possibilities.
You roll 4, 1; 3, 2; 2, 3; or 1, 4; or sum of 9.
That's the event I called E sub 2.
It happened to be exactly four possibilities for that
as well, either a 6, 3; a 5, 4, a 4, 5; or 3, 6.
So the probability of the sum is either 5 or 9
is simply how many ways you can get a 5 or a 9,
which happens to be 8, over the number of ways
you could get any sum, which as we know from earlier problems
is 36.
Web Assign will accept 836 as the answer.
Or you can reduce it to 2/9.
But there is something about those last two examples
worth noting.
There's a difference.
Look at those two.
I'm flashing on the left.
The first one I did.
On the right, the second one I did.
Going back to the one on the left,
the probability of the sum was 5 or 7.
It turned out just to be the sum of the probability
that the sum was 5 and the probability of the sum was 7.
Other words, there were four events that gave me a sum of 5.
So the probability of getting a sum of 5 is 436.
But there are also four that gave me a sum of 8.
So the probability of getting a sum of 8 is also 436.
And 436 plus 436 is 836.
But if you look on the right side,
the probability that a number is greater than 2
or odd turned out to not be equal to the probability
that the sum was 2 plus the probability of the sum was odd.
If you look at it, the probability of the sum
was 2, 3, 4, 5, 6, 7, 8, 9, 10 was 8 out of 10.
The probability of the sum was odd was 5 out of 10.
Well obviously, 8/10 plus 5/10 is 13/10 not 9/10.
So on the left, I was just able to take the two events
separately and add up the probabilities
and get the correct answer.
On the right, if I tried that trick it wouldn't work.
Why is that?
It's an important distinction.
Think about it just a minute.
In a nutshell, it's because on the right some of the values
are greater than 2 and odd.
Remember that when we did this problem on the right,
there were certain numbers that were both greater than 2
and odd.
For example, 3 is greater than 2 and odd.
5 is greater than 2 and odd.
7 is greater than 2 and odd.
And 9 is greater than 2 and odd.
So there's an overlap between those counts.
On the left, that didn't happen.
When you looked at the numbers that summed to 5
and the numbers that summed to 7, there was no overlap.
That's what caused the difference.
Noting this leads to an important result
called the addition or union rule for probabilities.
Remember, that's one of the ones that I flashed up earlier.
It says that the probability of E or F
is equal the probability of E plus the probability of F
minus the probability of E and F.
And if you look at that formula, it encapsulates the difference
that we just noticed between those two early examples.
You could have used this rule to solve
both of the earlier problems, but you don't have to.
We were able to figure out a way to do it
without using the formula.
But I will say that if you did use the formula
for the first one, where the advance were
mutually exclusive--
the sum being 5, there was no overlap with the sum being 7.
There was nothing to subtract away.
If you take that minus part of the formula
it would have been 0.
The probability would be 0 that a sum could be 5 and 7.
However, on the second one, there was an overlap.
There were numbers bigger than 2 that were also odd.
This formula takes care of that.
Think about that what that means.
If the events are mutually exclusive,
the probability of them both occurring together is 0.
So you're actually subtracting nothing.
So the formula also allows for the possibility
that that overlap is not probability 0.
So the formula works either way.
But if they're mutually exclusive,
remember, just nothing subtract.
The formula just degenerates to having to subtract nothing.
Now let's look at some problems where
we might want to think about it in terms of using this formula.
If we select a single card from a standard 52-card deck,
what is the probability that we draw either a heart or a face
card.
So the probability of a heart or a face card.
Recall with this standard 52-card deck looks like.
It has 4 suits, 12 face cards, et cetera, et cetera.
So the probability of a heart or a face card
is just going to be the number of hearts
or face cards over the number of cards in the deck.
Well, if you look at the hearts or face cards
and just count them, the face cards
are the cards that actually have faces on them.
And you've got the hearts as the other possibility.
If you just count, you'll see that there are 22 of them--
over the number of cards in the deck, which is 52.
So you get 22 over 52.
And if you want to reduce it, you get 11 over 26.
Suppose I wanted to actually use that addition rule,
rather than just figure it out as I just did here.
It would still work.
I would just simply take the probability of getting a heart.
Well remember, there are 13 hearts in the deck.
So the probability of getting a heart is 13 over 52.
Then I would calculate the probability
of getting a face card.
Well, there are 12 face cards in the deck--
4 jacks, 4 queens, 4 kings.
So the probability of getting a face card is 12 over 52.
And if I apply the formula, I'll see that I still
need to calculate the probability that a card is
both a heart and a face card.
And if you look at the picture, there
are three heart that are also face cards.
That's the jack of hearts, the queen of hearts,
and the king of hearts.
So the probability that the card is both is 3/52.
And if you plug in to the addition rule formula,
you'll notice that you get the probability of getting
a heart is 13/52 plus the probability of getting a face
card is 12/52 minus the probability
that you get a card that's a face card and a heart--
that's 3/52.
And that's 13 plus 12 minus 3 is 22.
So you get 22 over 52, which is 11/26, which is exactly
the answer we got earlier by doing it without using
the addition rule.
So this is just another example of where
the addition rule works.
But you could really do it without it.
Here's an example where using the rule
is really the best way to do it--
the most straightforward way to do it.
It says if the probability of A is 0.3, the probability of B
is 0.4, and the probability of A and B is 0.2.
Find the probability of A or B.
Remember, and is intersection, or is union.
So I would say the most straightforward way
to do this particular problem is to use the formula.
If you write the formula down, replacing E and F with and b,
you get the probability of A or B is equal the probability of A
plus the probability of B minus the probability of A and B.
Well, we know the probability of A is 0.3.
We know the probability of B is 0.4.
And we also know that the probability of A and B is 0.2.
So a little bit of arithmetic tells you that the final
answer-- the probability of A or B--
is 0.5.
That's the most straightforward way to do it.
Another example where the formula
is the most straightforward way to do it,
what if the probability of A is 0.3, the probability of B
is 0.4, the probability of A or B is 0.5.
But this time we're looking for the probability of A and B.
So it's the same formula, but you know different things.
Here, you know the probability of A or B is 0.5.
The probability of A is 0.3.
The probability of B is 0.4.
And you're looking for the probability of A and B.
It becomes a small algebra problem.
It's easy enough.
0.3 plus 0.4 is 0.7.
If you move the negative probability to the left
and move the 0.5 to the other side, making it negative,
you'll end up finding that the probability of A and B
is 0.7 minus 0.5, which is 0.2.
These are both examples where using the addition or union
rule for probabilities is the most straightforward way
to solve the problem.
Use the following table to find the probability
that a randomly chosen member of the student government board
is a freshman or lives in off-campus housing.
They give you a table.
This is another problem where you probably can just
do it by the old techniques.
You want to know the probability that a person is a freshman
or lives off campus.
So you look at your chart.
There are five freshmen and there are five students
who live off campus.
So we want to know the probability of being a freshman
or living off campus.
You can't just say 5 plus 5, because there's
one person who overlaps.
There's one person who is a freshman,
but also lives off campus.
And when you count the number of freshmen or students
living off campus, you have to say 4 plus 1 plus 4.
That's only 9.
And then when you count everybody,
you get 4 plus 1 plus 2 plus 0 plus 4 plus 2
plus 1 plus 0 plus 4 plus 0, 0.
You get 17.
So the answer is 9/17.
Again, you could use the addition rule
to get the answer.
But I would recommend just doing it straightforwardly.
Consider the experiment of rolling two dice.
Let A be the event of rolling a sum of 8.
And let B be the event of rolling a double.
A double means you get the same number on both dice, like a 2
and a 2 or 5 and a 5.
Find the probability of getting either a sum of A or a double.
Again, when you're rolling two dice,
you think of that as a two-stage experiment.
So you build the table.
We've talked about that before.
The probability of getting an 8 sum or a double
is going to be the number of ways of getting a sum of 8
or a double divided by the number of ways to get any sum.
So if you look at the numerator, let's
first look at the number of ways of getting a sum of 8.
You could get a 6, 2; a 5, 3; a 4, 4; a 3, 5; or a 2, 6.
Then look at the number of ways to get a double.
You could get a 1, 1; a 2, 2; a 3, 3; a 4, 4; a 5, 5 or a 6, 6.
And you want to count all those possibilities.
Notice, there's an overlap of 4, 4.
So when you're counting, you don't
want to count that 4, 4 twice.
If you count them, you get 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Don't count that 4, 4 twice.
So that gives you 10 over the number of ways to get any sum.
Well it's, as we've learned from doing these type
problems multiple times, there are 36 possibilities.
So the answer is 10 over 36.
And if you want to reduce it you get 5/18.
Again, we didn't have to explicitly use the addition
rule for probabilities, although you could.
Let's go to the complement rule that we
introduced in our toolbox at the beginning.
The complement of an event.
Let's take an experiment where we're
rolling a single ordinary dice.
The sample space obviously, is going to be either 1, 2, 3, 4,
5, or 6.
So if we want to look for the event
that we rolled a 3, the probability of doing
that is the number of ways you can
get a 3, which if you're rolling a single die,
there's only one way to do that, over the number of elements
in the sample space.
So the probability of rolling a 3, is 1/6.
The complement of that event--
and we're going to use the notation
E with the superscript c--
some text use E with an accent symbol
like we used back when we talked about regular set operations.
But if you see it either way, it means the same thing--
E with a superscript of c or E with an accent symbol.
It's sort of in a sense, the opposite of the event.
So if you're looking for an outcome,
the complement is everything except that outcome.
So if you're talking about the problem
we did above, if the event E was getting a 3,
then E complement is not getting a 3.
So in this case, if we wanted to write out E complement,
it would be everything except 3.
So it would be 1, 2, 4, 5, or 6.
So when we're talking about the complement event, that's
what we mean, everything but what makes up the event.
There are five things that aren't 3's out of possible 6.
So you get 5/6.
And notice that the probability of getting a 3 is 1/6,
and the probability of not getting a 3 is 5/6.
And that's not a coincidence.
That's what the complement rule says.
It simply says if you know the probability of an event,
and you want to know the probability of the complement
event, you just take 1 minus the probability of E.
And importance of that rule is-- it same sort of obvious--
but it's very useful because some problems,
it's really easy to calculate the probability of an event,
and it's a little more difficult or more than a little more
difficult sometimes to calculate directly, the probability
of the complement.
But you can calculate the probability event,
all you have to do is say 1 minus that probability,
and you'll automatically have the probability
of the complement event.
So this is important.
It's a very handy tool to have in your tool box.
For example, if I ask you to find
the probability of not rolling a sum of 11
when you're rolling two dice--
again, you'd build your table--
the probability of not rolling an 11
is the same thing as 1 minus the probability of rolling an 11.
The probability of the complement of something
is 1 minus the probability of that something.
So it's very easy to calculate the probability if sum is 11,
because there are only two possibilities.
You can get 6, 5 or 5, 6.
So the probability of getting a sum of 11
is simply 2 out of 36.
So the probability of not getting a sum of 11
is 1 minus 2 out of 36, which would be 34 out of 306.
Complement rule is a very powerful tool in your tool box.
If you wanted to reduce that to 17/18, you're free to.
I want to talk a little bit about the English language
as relates to this.
It turns out that when we're doing probabilities,
we tend to come to these problems
where we're asked to find the probability of at least
one something-- and I want to talk a little bit about that
in particular, so that when you see it, it won't be confusing.
For example, if I roll a standard die three times,
what's the probability we get at least one 4?
You see that at least one quite often.
At least one 4.
We'll think about it.
At least one 4 means you get exactly one 4,
or exactly two 4's or exactly three 4's.
That's what it means to have rolled at least one 4
in the context of rolling the die three times.
One way to answer this question is just
to do three separate probability calculations
and just add at the results.
Just find out the probability of getting exactly one 4.
Then find the probability of getting exactly two 4's.
Then find the probability of getting exactly three 4's.
And just add them all up.
That's one way to do it.
You could take this and this and this,
and just add up the three separate probabilities.
That's quite a bit of work.
There is a better way.
A better way is to use the complement rule,
because if we roll the die three times,
there are only four possibilities altogether.
We've talked about three of them.
We've talked about getting one 4 or two 4's or three 4's.
There's only one other way.
And that's to get no 4's at all.
So wouldn't it be much easier to calculate
the probability of not getting any fours
and then say 1 minus that?
And that's what I'm going to do.
The complement rule says the probability
of getting at least one 4 is 1 minus the probability
of getting no 4's, because getting no 4's is
the complementary event of getting at least one 4.
And getting no 4's is a much easier probability calculation.
So how did we calculate the probability of getting no 4's?
And this comes back to the stuff we talked about earlier.
Remember when we started studying this,
I asked you if you know how to count.
The solution that problem requires
us to count using some of our combinatorial formulas.
And that's why we studied them, so we could use them.
So if a standard die is rolled three times,
what is the probability that we get at least one 4?
Well, we just talked about the easiest way to do it is
to calculate the probability of getting at least one 4,
as 1 minus the probability that we don't get any 4's.
That probability of getting no 4's is the key calculation
in the whole problem, once you understand
what you're trying to do.
So how do you calculate the probability you don't get any
4's?
Well, I would think of it this way.
The probability of getting no 4's is the number
of ways to get anything but a 4 divided by the number of ways
to get anything with your three rolls.
I think I'll start with the denominator first.
Let's find the number of ways to get any three rolls.
Going all the way back to the counting principle,
if you're looking for the number of ways to get any three rolls,
remember each roll can be a 1, 2, 3, 4, 5, or 6.
Think of this as a three-stage experiment.
You can look at how many ways you can get something for roll
one, row two, roll three.
The counting principle says to multiply them together.
So the number of possibilities for the first roll-
well, there are six possibilities.
As for the second, there are still six possibilities.
And for the third, there are still six possibilities.
So if the counting principle says the possible die rolls
you can get when you roll three times is simply 6 times 6 times
6, which happens to be 216.
That's the denominator.
But what about the number of ways to get anything but a 4?
Well, after having done the calculation we just did,
I think you'll see that it's pretty
easy to look at the numerator and see that it's
the same thought process.
But this time we're not allowing the possibility
of getting a 4, because we're looking for the number of ways
to get anything but a 4.
So we knocked the 4 out as being a possibility,
but the process stays the same.
We're still doing three die rolls.
We just don't want any 4's.
So there's only five possibilities
for the first die roll if you don't want it to be a 4.
There are only five possibilities for the die roll
two if you don't want it to be a 4.
And there are only five possibilities
for die roll three if you don't want it to be a 4.
5 times 5 times 5 is 125.
So the probability of getting no 4's is simply 125 over 216.
But that wasn't the final answer.
We want to know the probability of getting at least one 4.
The key calculation was getting the probability of no fours,
but it wasn't the final answer.
The probability is 1 minus that, which would be 91 over 216.
How about this?
If you draw three cards from a standard deck
without replacement, what is the probability that at least one
is a face card?
Write your answers as a percent rounded to one decimal place.
So as we learned, the probability
of at least one something is 1 minus the probability
of none of that thing.
So in particular, the probability
of at least one face card is 1 minus the probability
of no face cards, or zero face cards.
And as we've seen before, that probability of zero face cards
is the key calculation.
Well, what is the probability of no face cards?
There are 12 face cards in the deck.
We've done this multiple times.
So that means there are 40 cards that aren't face cards,
because 52 minus 12 is 40.
So we think of partitioning the deck into face
cards and non-face cards.
There are 12 face cards.
There 40 non-face cards.
So the probability you don't get a face card
is the number of ways not to get a face card--
in other words, the number of ways
to get anything but a face card--
divided by the number of ways to get any three cards.
You've done these problems in earlier sections.
When we're drawing face cards, we don't care about order.
Because when you pull a card out of the deck,
you put it in your hand, it doesn't matter what order
the card came into your hand.
So this is a combination.
If we're trying not to get a face card,
we've got to draw from the 40 things that aren't face cards.
So it's a combination of 40 things,
and we're choosing three cards.
So it's a combination of 40 things, taken three at a time.
Down bottom, when we don't care what the cards are,
it can be any 52.
But we're still drawing three cards.
So it's a combination of 40 things,
taking three at a time-- that's the number of ways to get
anything but a face card--
over a combination of 52 things taken three at a time.
That's when you don't care what the cards are.
You're drawing three.
But you don't care what they are.
And in case you really have forgotten your shift
combinations to get your calculator answer,
I put them there for you.
I hope you didn't need it.
They ask you to do a decimal approximation.
So you need to change that now to a decimal.
That's 0.44706.
They ask you to change it to a percent rounded
to one decimal place.
So take plenty of decimals, because you have to change it
to a percent and then round it.
So do not just write 0.44.
Write several decimal places.
So it's 0.44706 to several decimal places.
So you continue your calculation.
The probability of getting at least one face card
is 1 minus the probability of no face cards, which
we've calculated to be 0.44706.
Notice I'm emphasizing yet again,
carry plenty of decimal places during this calculation.
Round off in the very last step.
So do that subtraction.
You get 0.55294.
You're still not ready to round, because it
says to give your answer as a percent rounded
to one decimal place.
So you first change it to percent, and then you round.
You sort of see now why I'm emphasizing
how important it is to carry plenty of decimal places.
Now change it to percent, which means
you move the decimal place two places to the right.
And finally, you round that to one decimal place.
Suppose we take the problem that we just solved
and change it ever so slightly so that now we're
drawing three cards from a standard deck,
but we're drawing with replacement.
We still want to know what is the probability that at least
one is a face card.
And we'll take our answer as a percent
rounded to one decimal place.
So the only difference here is that we're
drawing with replacement.
How does that change everything?
What's different about it?
Because we're drawing with replacement,
this does not involve counting things
using either the permutation or combination formulas.
Those are what we use when we have distinct objects,
and we're looking at how many arrangements--
or how many ways we can choose from those.
When we're drawing with replacement,
it just doesn't fit that.
So we have to go back to the counting principle for problems
like this.
And that's the difference between this problem
and the one you just solved.
But still, with the complement rule applies--
and we can calculate the probability
there's at least one face card by calculating 1
minus the probability that there aren't any face cards,
or that there is zero face cards.
So this part doesn't change at all.
And that calculation of getting zero face cards
is really the key to the whole thing.
Because once we get that, the final answer will be just 1
minus that answer.
Remember also, there are 12 face cards.
So if we're trying to not draw a face card,
we're looking for one of the non-face cards.
And there are 40 of them, because 52 minus 12 is 40.
And as before, the number of ways you can get no face cards
is simply the number of ways that you don't get a face card
when you're drawing those three card over any possible way you
could draw three cards from the deck.
And again, this time with replacement.
So the answer for the zero face cards probability
will be a fraction.
And we have to calculate the numerator and denominator,
and we're home free.
But as I said, when we're doing it with replacement,
we can't calculate it with our combination formulas.
We have to go back to the counting principle.
And here it says we're just doing
three events, three sub-tasks-- however
you want to think of it.
And we've got to count the number
of ways you can do each one.
Well, if you're trying to not draw face card,
on the first draw, there are 40 ways
of not getting a face card.
And since you're putting the card back in the deck,
there are 40 ways to not choose a face card on the second draw.
And again, because you're putting the card back,
there are 40 ways not to choose a face card the third time.
So the numerator will be 40 times 40 times 40.
How about the denominator?
Well, same thought process, but in the denominator,
you don't care what it comes out to be.
It could be any of the 52 cards.
And again, because you're putting it back in,
you still have 52 cards to choose from.
You don't care what the card is.
And the third draw is the same.
You have all 52 cards to choose from.
And you don't care what the card is.
So the result will be 40 times 40 times 40 over 52 times
52 times 52.
That's 64,000 over 140,608.
Now, the answer they want is decimals rounded to one place.
And that's a percent.
So we need to change that fraction
to a decimal at some point.
You might as well do it now.
Always take quite a few more decimal places
than you think you're going to need.
It says the round it to one decimal place.
But it also says to write as a percent.
Writing it as a percent causes you to move the decimal place
two places to the right.
Then you want another decimal place of accuracy.
That's three.
So definitely take, I would say, five or six decimal places,
for sure.
So I wrote it as 0.455166.
Now remember, the probability of at least one face card
is 1 minus the probability of no face cards.
And we just calculate the probability of no face cards.
So 1 minus that value is 0.544834.
Now we're done, except for the fact
they asked us to do it as a percent rounded
to one decimal place.
So changing it to a percent moves the decimal place
two places to the right.
And then you can see that that first decimal place is a 4.
But the hundreds places is an 8.
So that 4 rounds up to a 5.
So as a percent rounded to one decimal place,
it would be 54.5%.
So if you draw three cards in a standard deck with replacement,
the probability that at least one is a face card
is over 50%, 54.% approximately.
Now in the next section, we'll learn another method
of solving this type of problem that you
might find a little easier.
But for now, this will get you there.
Now that I have a problem like this,
I'm trying to give you variations of problems that,
on the surface, sound very similar,
but may have a different technique of solving them,
because other techniques you've been using don't apply.
How about this?
A pair of dice is rolled three times.
What is the probability that we get a sum of 6 at least once?
And this time, they're not asking for percent answer,
but they do want the final decimal rounded
to three decimal places.
We worked a problem similar to this
before, except that we didn't roll a pair of dice.
We just roll one die.
But I didn't want to go back and remind you
of all those similarities.
Not only is it similar to that one,
but it's also very similar the one we just worked with cards.
The difference there is that we're
doing probabilities based on two dice
here instead of drawing cards from a deck.
And as I said, it's extremely similar to one
we did somewhat earlier in this lecture
where we were rolling just a single die, but otherwise,
very similar.
So I put up in the corner a screen capture
from that calculation.
And you can see what we did.
And we just did it a few minutes ago.
So it should be fresh in your mind.
The point being, neither the problems involving
die rolls, either this one with two dice or the other one
involving one die, which are independent events-- rolling
a pair of dice once and rolling them again,
those events are completely independent of each other.
The probabilities don't change because of a previous roll--
nor the previous card problem where we drew with replacement.
Because we were drawing with replacement,
that made those draws independent.
None of the problems involve counting things
because either permutation or combination formulas.
We can't use those when we're drawing
with replacement or any case where
there are independent events.
So again, we've got to go back to the counting
principle for this solution.
But otherwise, it's very similar to the one die roll problem.
If you look at that in the lower right corner,
the only difference is instead of looking at probabilities
involving getting fours, which is what the other problems are.
We're trying not to get a sum of 6.
Now any time you say the sum of rolling two dice,
you've got to remember that we've
got to go back and look at the sample space
for rolling two dice.
There are 36 possibilities.
We've done this multiple times.
But just to emphasize this, in the lower right hand corner,
we went from looking for the probability of getting
any fours when we're rolling a single die to what we really
want now, which is the probability involving
sums of 6.
That takes us to the sample space in the upper right hand
corner, which is a 36-member or 36-element sample space.
Once we get this, all we have to do
is figure out the value that goes in the numerator
and denominator.
And I think maybe it's easier to start with the denominator.
So let's look at the denominator first.
Using the counting principle, remember,
I said we'd have to go back to the counting
principle for these problems where
the events are independent.
We're going to do three roll of a pair of dice.
On the first roll in the denominator
we don't care what the sum is.
So it could be any of those 36 sums.
So you've got 36 possibilities for the first row,
36 possibilities for the second row,
36 possibilities for the third row.
And 36 times 36 times 36, by the counting principle, is 46,656.
That's the denominator of our answer for the probability
of getting no sums of 6.
Let's move to the numerator.
Now the numerator, we don't want any sums of 6.
We're looking for the probability
of getting no sums of 6.
Well, if you look up there, there are 31 of them.
5, 1 adds to 6.
4, 2, adds to 6.
3, 3 adds to 6.
2, 4 adds to 6.
1, 5 adds to 6.
In other words, only that diagonal line
of sums that I did not circle add to 6.
We want the ones that don't add to 6.
And if you count those circles or just say 36 minus 5
that weren't circled, you'll see that there
are 31 possibilities.
And again, this is a counting principal problem
and we've got three rolls.
On the first roll, any of those 31 that I circled are OK.
Yet none of those 31 give us sum of 6.
Again, you roll a second try.
Same thing.
31 possibilities.
Third roll again, there are 31 possibilities.
All you're screening for are sums that are not equal to 6.
And by the counting principle, 31 times 31 times
31 is the total number.
It comes out to be 29,791.
And that answer goes in the numerator.
So if you take your calculator, you get about 0.63852.
Again, take plenty more decimals than you need,
because you want to make sure you're rounding error
doesn't cause you to miss the problem.
Also remember, we're not looking for the answer, not
the final answer, as being the probability
that there are no sums of 6.
We're actually looking for the probability
that you get at least one sum of 6.
But we know by the complement rule
that the probability of at least one sum of 6
will be 1 minus the probability of no sums of 6.
We just calculated the probability of no sums of 6
being the 0.63852.
So if plug that in and subtract 1 minus that,
you get zero 0.36148.
They ask us not to change it to a percent this time,
but just to round it to three decimal places.
That would be 0.361 rounded to one decimal place.
So the probability of rolling a pair of dice three times
and getting at least one sum of 6 is 0.361.
In wrapping up this lecture, I do want to say one thing.
When you're practicing and working
through these problems and similar problems,
try looking at the big picture.
If you're isolating each problem as a separate thing
and just concentrating on solving that problem as you
move from problem to problem, you're probably
not seeing the big picture.
And when you come back to those kinds of problems on the test,
you may have trouble solving them.
Now I'm going to mention a few things.
But this is not an exhaustive list.
But for one thing I mean by that just the stuff that we've
been dealing with recently about whether you're drawing cards
with replacement or without replace replacement, when
you're rolling die--
one die, two dice--
you've got to be careful.
If the things involve independent events
like rolling dice or drawing cards with replacement,
you're generally going back to the counting principle
because the combination and permutation formulas are
generally used to count arrangements
or choosing objects from distinct objects.
And that simply doesn't imply when
you're drawing with replacement or rolling dice.
That's the first thing.
You need to notice when you're reading the problem is
it an independent event--
rolling dice, drawing with replacement.
Or is it a dependent event?
If it's a dependent event, when you're drawing cards
from a deck and not putting them back,
generally you'll be able to count using your combination
formulas.
So that's one big picture item.
Another thing I've mentioned in closing
is that the usual strategy when you computed
the probability of at least one of something,
you handle that by applying the complement rule.
In other words, go ahead and calculate
the probability of getting none of those things.
And then the probability at least one of them
is 1 minus the probability of none of them.
There are other things I could mention,
but I just want to get you thinking
about looking for the big picture
as you work through these problems.
It's going to make your test taking experience much better.
Section 12.5 on conditional probability--
I want to talk about the idea of independent and dependent
events.
I alluded to that in the earlier formulas
when I put up those formulas in advance.
Independent events are ones for which the result of one event
does not affect the probability of the occurrence of the other.
There are some simple examples that we'll go through.
But I need to throw up a thing to contrast it with,
which are dependent events.
Dependent events are one for which
the occurrence of one event affects the probability
of the occurrence of the other.
Now that you can see the contrast,
I want I put some examples up and see
if I can make it even clearer.
Let's decide whether the following events
are dependent or independent.
Doing your assignment and passing your math class.
Do you think doing your assignments might
have some effect on the probability of you
passing your math class?
Yeah, I think most people would agree to that.
So those are dependent events.
One thing can affect the probability of the other thing.
How about this?
Running daily and winning the New York Marathon?
Again, I think most people would agree
that that might have an effect, that daily running might very
well have an effect on how probably it
is that that person will win the New York Marathon.
So I'd say again, those are dependent events.
How about this?
Winning the lottery and winning a spelling bee?
It's really hard to imagine how winning a lottery
could affect the probability of winning a spelling bee,
or the probability of winning a spelling bee
would have any effect on the probability
of winning a lottery.
So those two events are independent.
Getting heads on a coin toss and rolling a 6 on the standard
die.
Flip a coin.
Does it matter what you get in terms
of it changing the probability you're going to roll a 6
on the standard die?
No, I don't think so.
So those are also independent events.
This sort of gives you an example
of what we mean by dependent and independent.
You may only need an intuitive idea of what they mean.
How about this?
Drawing an ace, then drawing a 2 from a deck of cards
without replacement?
Now think about this.
If you draw an ace out, you don't replace it,
and then you draw a 2, that does affect the probability,
because you since you didn't put that ace back in,
you're only drawing from 51 cards the second time.
There is a dependency there.
Drawing that ace out and not putting it back in
makes those two events dependent.
If you put the ace back in before you drew again,
then that would erase the dependents.
And they would be independent events.
These are very important distinctions.
You need basically an intuitive idea.
But having an intuitive idea is extremely important.
And this leads to something called the product
rule for independent events.
If two events E and F are independent,
then the probability of E and F is the probability of E
times the probability of F. That was one of the tools
that I flashed up at the very beginning,
and said we'd get to it later.
Suppose you flip two coins.
What is the probability that the first clip is heads
and the second flip is tail?
The coin flips are independent.
What I get the first time I flip the coin has absolutely nothing
to do with what I get the second time I flip the coin.
So that means they're independent.
So that tells me that the probability
I get a head and then a tail is simply
the probability I get ahead times the probability I get
a tail, because that's what the product rule
for independent events says.
If the probably I get ahead is 1/2.
The probability I get a tail is 1/2.
So the probability that I get a head followed by a tail
is 1/4, a 1/2 times 1/2.
That's the product rule for independent events.
I just want to say that you could have worked
this problem with a tree diagram if you look at your first toss
as being the first stage of a two-stage experiment--
the probability of getting a head is 1/2.
And then once you've done a head,
the probability of getting another head is 1/2,
and the probability of getting a tail is also 1/2.
So if you look at the branch that goes from heads to tails,
you see it's 1/2 probability you get ahead,
and there's one 1/2 probability you get a tail.
And we learned earlier that the probability multiplying down
a branch is the and probability. so the probability
of getting a head and a tail, where the first is
a head and the second is tail, it would be 1/2 times
1/2, which is 1/4.
So we did learn something in earlier sections
that would allow you to get that same answer with a tree
diagram.
But treating it as just two independent events
is much simpler.
Now I want to talk about conditional events.
That is the subject of section 12.5.
And this is also where I said earlier,
we introduced a new notation.
The conditional probability of event E occurring
given that event F has occurred is the probability
of the event E occurring, taking into account
that event F has already occurred.
And that's what the adjective conditional means.
Conditional probability means the probability
conditioned on the fact that something else has already
occurred.
It's good to probably introduce this idea through an example.
Let's let E be the event that the sum is
5 when we're rolling two dice.
What is the probability of E?
Well, that's just the probability we get a sum of 5.
And we know how to do that.
We draw our 2 by 2 table.
And we count the sums that come to 5, which happened to be 4.
And we get, by the basic probability principle
is the number of ways of getting what you're looking for,
which in this case is the sum of 5 divided by the number of ways
you can get anything in the sample space, which
we know there 36 things.
So it would be 4 over 36.
If you want to reduce it to 1/9, you can.
But what if I tell you that I've already peaked?
I've done the die roll already.
And I'm holding my hand so you can't see it.
But I've peaked.
And I'm telling you something about it.
I'm not telling you what the sum is,
but I'm going to tell you-- well,
I'm not telling you what it is, but it's odd.
Does that matter?
Does that change the probabilities
because I gave you some new information?
And the answer is in general, yes,
that will make a difference.
So what I really want now is the probability
of the event E happens, which is getting a sum of 5,
given that additional information, given
that the sum is odd.
That additional information is important.
Because if I know the sum is odd,
that changes the sample space, because I
don't have to consider any even sums anymore.
So I can get rid of every sum that comes out even.
1 plus 1 is 2.
That's even.
5 plus 3 is 8.
That's even.
5 plus 5 is 10.
That's even.
6 plus 6 is 12.
That's even.
So I can get rid of all of those.
So that additional information is extremely useful.
I can throw out the things that no longer contend.
Because I know it's odd, the evens don't content anymore.
So the probability of E given that the sum is odd
is just the number of ways that E
can happen with that new condition-- in this case,
that the sum is odd-- divided by how the samples place changes
given that new condition.
You're still doing a count of the event E.
And you're still doing a count of event S,
but you're taking into consideration that something
has happened and you have more information
than you had to start with.
That's what I mean when I say with condition.
Well, once I crossed out all the even sums, there were 36.
Half of the sums were even.
That knocked the sample space down to 18.
There are only 18 odd sums.
And then I look--
none of these sums adding to 5 got knocked out,
because they were all odd.
So there are still four of those.
So the probability of the sum being 5,
if I know that the sum was odd, is going to be 4/18.
It would have been 4/26 had I not known that.
So instead of the probability of the sum being
5 being 1/9, now the probability that the sum is
5 with that additional knowledge--
that knowledge being that the sum was odd--
now has doubled to 2/9.
Yes, it does matter.
Or at least it can matter.
You can see here the probability doubled.
Knowing that information doubled the probability.
So to sum up, when we have a probability which
includes an extra condition--
and in this problem, it was that the sum was odd--
that probability is called a conditional probability.
And we have a special notation for it.
And if we go back to that previous example,
the probability of E--
I just wrote this sort of intuitively--
the probability of E comma given that the sum is odd--
we're going to modify that.
I just wrote that out kind of intuitively.
Suppose we call that condition F. In other words,
the sum being odd is a new condition called F.
So the sum being 5 was the event E. Now
we're going to call the some sum odd, the condition F,
then we'd write the conditional probability
with this vertical line.
The probability of E and the vertical line
is read as given that.
So that's the probability of E given F. So that new notation,
the vertical line is read given.
So this is the probability of E given F, or given
that F has occurred.
Probability of E given F. That's conditional probability.
This is important.
So take a minute to soak it in.
The probability of E given F is a conditional probability
of event E given that F has occurred,
that F is the additional information that we now know
that we didn't know before.
The vertical line is read given or given that,
depending on exactly how you want to phrase it.
For our particular problem, this would be the probability
that the sum is 5 given that the sum is odd.
Think about that a minute, because you
need to understand what the probability is saying or asking
before you can get an answer.
And if I don't make any other point in this lecture,
I want to say this and make this point.
Oftentimes, if you understand how the given that condition
restricts the sample space--
the original sample space-- you can
solve this conditional probability problem
without even formally using formulas.
But if you really understand how the given that condition
restricts the original sample space,
quite often you don't really even need formulas.
For example, if I told you that two cards are drawn
without replacement from an ordinary deck of cards,
and asked you the probability that the second card
is a heart, given that the first card was a heart--
They're asking me for the probability
the second card is a heart, given
that the first was a heart.
Well, if I've noticed the first was a heart--
I'm given the first card was a heart, then
that reduces my sample space.
If I know the first card was a heart, when I draw again,
instead of having 52 cards in the deck,
I'll only have 51 cards in the deck.
And since the first card was a heart
and there are 13 hearts in the deck,
if the first card really was a heart, after I draw a heart
out, I only have 12 left.
So knowing that the first card was a heart,
if I didn't put the card back, and it
says drawing without replacement, when I draw again
or just before I draw again, I'll
be drawing from a 51-card deck that only has 12 heart.
So now the probability of that second card being a heart
is simply 12 out of 51.
Knowing that the first card was a heart
and it wasn't re-replaced, reduced the sample space down
to 51 cards, 12 of them being hearts.
And that probability would be 12 out of 51.
Simple, if you know how to think about it.
And if you want to reduce that to 4/17, you can.
How about this one?
A container has 35 green marbles, 20 blue,
and 4 red marbles.
Two marbles are randomly selected without replacement.
If E is the event that a green marble is selected first,
and F is the event that the second marble is not green,
compute the probability of F given E.
I will say that when I say these problems where the events are
given letter names, if they tell me what the actual events are,
my inclination is to go ahead and replace
the letters with an actual phrase representing the event.
So if F is the event that the second is not green,
I'll just write that out.
And if E is the probability that the first was green,
I would just write that out.
It's up to you.
But I think it's easier if you do it that way.
Now having done that, remember we're
given that the first draw resulted in a green marble.
And we're doing it without replacement.
So if there started off being 59 marbles--
after the first draw, there are only 58 marbles.
And the one that was drawn was a green.
So instead of 35 there are 34 green.
We want to calculate the probability
that the second was not green, given the first was green.
You can see that I've already put the answer up there
as being 24 over 58.
And I think you can see that where that came from.
The denominator is simply the count
of how many marbles were left.
And the numerator are the number of things
that are not green, the number of marbles not green.
And you can see is there are 20 blue and 4 red.
And 20 plus 4 is 24.
So there were 24 non-greens to choose from.
So the probability is 24 over 58 non-greens
over total marbles left.
And it's 58 instead of 59, because we're were
drawing without replacement.
If you want to reduce that, you can reduce it to 12/29,
but Web Assign really doesn't care.
How about this one?
Two dice are rolled.
What is the probability that the sum of the dice
is 6, given that the row was a double?
Again, you want your sample space in a table, 2 by 2.
Six possibilities for the first roll.
Six possibilities for the second roll.
You're given that the roll was a double.
So we went the sum to be 6 with the additional information
that we rolled a double.
So the probability that the sum is 6, given
that the roll was a double.
Well, if we know the roll is a double,
we know that's already occurred.
We can get rid of everything but the doubles.
So the doubles are 1, 1; 2, 2; 3, 3; 4, 4; 5, 5; 6, 6.
So all of a sudden, that 36-member sample space
reduced down to 6, because we know that it was a double.
So we'll be dividing by 6.
Out of those remaining 6 choices,
only one has a sum of 6.
And that's 3, 3.
3 plus 3 is 6.
So the probability is 1 out of 6 or 1/6.
See how this works?
If you can think it through, you don't really
have to memorize the definitions.
You just have to understand what the condition does
to the sample space.
How about this one?
Mrs. Fraga's class has 109 students
classified by year and gender.
It's listed in the table below.
She randomly chooses one student to collect homework.
What's the probability of selecting a female, given
that she chooses a random--
that she chooses randomly from only the sophomores?
So the given that part is that she's only
choosing from the sophomores.
That means you can eliminate all the freshmen, all the juniors,
all the seniors.
So your sample space has reduced down to just the 21 sophomores.
So now what's the probability of selecting a female?
We'll look within the remaining choices, those 20 mining
choices.
5 of them are female out of the 21 sophomores.
Notice again, I said it.
But I'll actually write it.
There are 21 sophomores.
So the probability that the person--
the student was female, given that the person was
a sophomore is going to be the number of females
over the remaining things in the sample space, which are just 21
sophomores.
So your answer is 5/21.
Very simple if you understand what you're trying to do.
This might be a good place to mention
that we can use the conditional probability ideas to modify
the product rule for independent events
that we've talked about earlier to handle cases
where the events are dependent, as we've been talking about.
Recall that the product rule for independent events
says if two events E and F are independent,
that the probability of E and F is
equal to the probability of E times the probability of F.
But if the two events are dependent,
we can modify that slightly to say that the probability of E
and F is equal to the probability of E times
the probability of F given E. It's still a product,
but you're using that conditional probability.
The two events E and F are dependent.
Then the probability E and F is the probability of E
times the probability of F, given E. Let's do an example.
Let's draw two cards from a standard deck of 52
without replacement.
What's the probability that your first card is a spade
and your second car is red?
Recall your deck of cards.
You have to know enough about a deck of cards
to answer these sort of questions.
The first card being a spade causes
you to look at the deck of cards and realize that there
are 13 spades in the deck.
And the second card being red causes
you to look at the deck of cards and realize that 26 of those 52
cards are red.
The product rule for dependent event
says if two event E and F are dependent,
then the probability of E and F is the probability of E times
the probability of F, given E. Here,
that would be the probability of the first being
a spade and the second being a red is equal to the probability
that the first is a spade times of probability
that the second is red, given that the first is a spade.
And that should be easy enough to figure out.
What is the probability that the first is a spade?
Well, as we said, there are 13 spades in the deck.
So the probability of drawing a spade on that first card
is 13/52.
Also, keep in mind, you could reduce 13/52 to 1/4
if you want to.
You don't necessarily have to, but you can.
What about the probability of the second one
being red, given that the first was a spade?
This is a little tricky if you don't think because you're
drawing without replacement, which
means if you draw out a spade, you're not putting it back in.
So the deck now only has 12 spades and one less card.
So that would be instead of 52 cards in the deck,
there would only 51 in a deck.
There are still 26 red cards.
But now one of the cards has been taken out
and not replaced.
So there are only 51 cards left in the deck.
So the probability of getting a second red, given
that you've taken a spade--
it's not 26 out of 52, but instead is 26 out of 51.
And if you multiply 1 times 26 and 4 times 51,
you'll get 26 over 204.
And as I said, you could have left that 1/4 are 13/52.
You did not have to simplify it.
But it's nicer if you notice those sorts of things,
I believe.
So the answer would be 13 over 102,
if you simplified as much as you can.
Web Assign doesn't really care if you simplify fractions.
But if you want to, you can.
How about this one?
Three cards are down from a well-shuffled standard deck
of 52.
What is the probability that the first card is red,
the second card is black, and the third is red?
And give your answer as a fraction.
Remember, 52 cards in the deck.
The deck of cards has heart, diamonds, spades, and clubs.
There are 2's, 3's, 4's, 5's, 6's, 7's, 8's, 9's, and 10's.
And then there's jacks, queens, kings, which are the face card.
And then there aces--
52 cards.
The product rule says that the probability that the first was
red, and the second one's black, and third one's red,
is the probability the first was red times the probability
that the second one's black, given that the first one was
red, and times the probability the third one is red,
given that the first two choices were made.
So those last two are conditional probabilities.
It's basically the probability of getting
a red, the probability of getting
a black, the probability of getting another red.
But you're taking into consideration
what's already been drawn out.
Because if you're dealing cards, they're not being put back in.
So they're not going back in.
So that means it's without replacement.
That's what makes the conditional probability.
So anyway, we want to find out what is the probability
of going that first card red?
Well, there are 26 cards in the deck of 52.
So that's 26 over 52.
That also happens to be 1/2.
So if you wanted to write it is 1/2, that's perfectly OK.
I'll leave it that way for the time being.
But what about drawing a second black?
Now you've got to be careful.
After that first card is dealt, it doesn't go back in.
So there are only 51 cards left in the deck.
I drew out a red.
So there are only 25 reds and 26 blacks.
So the probability of getting a second black from that deck
of 51 cards is 26 out of 51.
I wanted black.
And there are still 26 blacks in there.
But there are only 51 cards in the deck.
So the probability of getting a second black,
given that the first one red is red, is 26 out of 51.
And finally, what is the probability
the third one is red, given that the other two were chosen?
Well, now you've taken two cards out the deck.
So there are only 50 cards.
And you've taken one of each color, one red, one black.
So there only 25 reds and 25 blacks left in the deck.
So the probability that the next one is red is 25 out of 50.
And according to the product rule, all you gotta do
is multiply those together.
Given your answer is a fraction-- if you just
multiply straight across, you'll get 16,900 divided by 132,600.
Now you can simplify that any way you want to.
In fact, you could have gone back up
to the top and simplified 26 out of 52 as being a half,
and 25 out of 50 being a half.
And you could have gotten a lot nicer answer.
But as said, Web Assign really doesn't care.
But if you did want to simplify, you
could get it all the way down to 13 over 102.
I want to wrap up this section by going back to three problems
that I covered in section 12.4.
And at the time, I said that I would show you
what I consider to be an easier way
to solve those three problems.
So let's do that.
If you look in the upper right hand corner of the screen,
you'll see I brought up a screen capture of that problem that
I'm talking about-- at least the first one of those--
and how we solved it before.
That problem said if we roll a standard die
three times, what is the probability that we
get at least one 4.
Instead of solving it the way I did in section 4,
I want to solve it using the complement rule, which
we did use in section 4, combined with the product
rule for probability that we covered in this section.
I think you'll find this much easier.
To do that, I start off with a complement rule just at the 4.
In this case, the probability of getting at least one 4 is 1
minus the probability of not getting any 4's or getting zero
4's.
Now the probability of getting zero 4's
is the key to this whole calculation.
Because once we get that, we just say 1 minus that,
and we'll have the answer we're looking for.
But the probability of getting no 4's, according
to the product rule of probability,
is simply the probability of getting no 4's
on roll one times the probability of getting no 4
on roll two times the probability of getting no 4
on roll three.
That's what we learned in this section.
But each of those probabilities is easy to calculate,
because if you don't want to get a 4,
you just have to get anything but a 4.
And also, because you're rolling die, the rolls are independent.
So the probabilities don't change.
So those will be three probabilities
that are exactly identical multiplied together.
Well, what is the probability of not getting a 4 on roll one?
Think about it.
If you're not going to get a 4 on roll one,
you can get either a 1, 2, 3, 5, or 6.
That's five possibilities out of six possibilities.
So the probability of that happening is 5/6.
And because the rolls are independent,
it's going to be 5/6 on the second roll
and on the third roll.
So the probability of getting no 4's
on any of those three rolls, is simply 5/6 times 5/6 times
5/6, which is 125 over 216.
Remember, we want to know the probability
of getting at least one 4.
So we have to take that and subtract it from 1.
So you have 1 minus 125 over 216,
which leaves 91 out of 216 as your final answer.
And if you look up in the corner of the screen on the right,
you'll see that's exactly the answer we got before.
I think this solution is much easier.
Let's do another one.
Up in the upper right hand corner,
I put up the solution that we used
to solve this problem in 12.4.
And you can see the answer in the solution technique.
I'm going to do the same thing with this problem.
This time we're drawing cards.
We're drawing three cards from standard deck
without replacement.
And we want to know the probability of getting at least
one that's a face card.
We'll start off just before using the complement rule.
But I want to do the rest of the problem using the product
rule for probability that we learned in this section.
So the complement rule here says the probability
of getting at least one face card is 1 minus the probability
that we don't getting face cards.
But remember, we're drawing three times.
And we're drawing without replacement.
So keep that in the back of your mind.
The probability of getting zero face cards on three
draws by the product rule is the probability
we don't get a face card in the first draw times of probability
we don't get a face card on a second draw
times the probability we don't get a face
card on the third draw.
But because these draws are without replacement,
these are conditional probabilities.
And they're going to change between draws because you're
not putting the card back in.
So keep that in mind.
What is the probability of not getting a face
card on the first card?
Well, if you look back in the upper right hand corner,
you'll see that we talked about this.
There are 52 cards in the deck.
There are 12 face cards.
There are 40 cards that are not face cards.
So the probability of not getting a face card
on the first draw is simply 40 out of 52.
But when we go to calculate the second probability,
the probability we dong get a face card on a second draw,
this is a conditional probability
because we're not putting the card back in.
So we got a non-face card on the first draws.
So that leaves only 39 of them out of 51,
because we're not putting the card back in.
So the probability on the second draw
of not getting a face card--
well, there are only 39 face cards left out of 51 cards.
And then you can see the pattern.
By the time you get to the third face card,
you're down to 38 face cards in the deck
with only 50 cards in it.
So the probability of getting no face cards by the product rule
is simply the product of 40/52 times 39/51 times
38/50, which comes out to be 59,280 over 132,600.
Now this particular problem ask us to calculate and leave
the answer as a percent rounded to one decimal place.
So we're going to have to change this to a decimal.
So this is probably a good time to do it.
You really have the probability of getting at least one face
card being 1 minus the probability
that you don't get any face cards.
And we just calculate that to be 59,280 over 132,600.
Probably a good time to change that fraction to a decimal.
We're going to change the answer to a percent,
and then we're going to round it.
So carry plenty of decimal places.
And then when you round later, you
will be accurate to one decimal place as a percent.
So carry plenty of decimal places.
And I carried it to five.
So when I divided 59,280 about 132,600,
I got 0.44706 to five decimal places.
1 minus that is about 0.55294.
But remember, the problem said to change it to a percent,
and then round it to one decimal place.
Changing to a percent involves moving the decimal place
two places to the right.
So that become 55.294%.
And now we want to round it to one decimal place.
So the 0.2 becomes 0.3.
And if you'll look up into the upper right hand corner,
you'll see it's exactly the answer we got before.
And I would contend that this solution is much easier.
Let's do one more.
From the earlier section, we did a problem
about rolling a pair dice three times,
and finding the probability that we get a sum of 6
at least once.
And again, we're going to round our answer
to three decimal places here.
We'll start off with the complement rule,
just as before back in the previous section.
But again, I want to use the product rule for probability
to solve this problem instead of the section 4 technique.
So according to complement rule, the probability
of getting at least one sum of 6 is
1 minus the probability of getting no sums of 6.
So the key to this whole problem is finding the probability
of getting no sums of 6.
According to the product rule, that
will just be the probability of not getting a sum of 6
in the first roll, not getting a sum of 6 on the second roll,
and not getting a sum of 6 on the third roll.
Remember, we're rolling two dice.
And you've got to remember when you're rolling two dice,
you go back to your sample space.
There are 36 possibilities.
And we've been putting that in a table.
So 36 possibilities-- and if you don't want a sum of 6,
you can see where I've circled all the sums that are not 6.
And the only ones that aren't circled
are 5, 1; 4, 2; 3, 3; 2, 4; and 1, 5.
That's five.
But there are 36 together.
That means there are 31 that are not
sums of 6's over a possible 36 sums,
irregardless of whether it's a 6 or not.
So the probability of getting no sum of 6 is 31 over 36.
And as in the problem before last,
these rolls are independent.
Rolling something on a first roll of those two dice
has no effect on the sum you get when you roll again.
Those are independent events.
So the probabilities don't change.
So it's going to be 31 over 36 times another 31 over 36
times a third 31 over 36.
And if you multiply that out, you'll get 29,791 over 46,656.
So the probability of getting at least one sum of 6
is 1 minus that fraction.
Again, they're asking us to leave the answer
as a decimal rounded to three decimal places.
So we want to change that to a decimal.
And carry plenty of decimal places
so you don't get a rounding error.
That comes out to be about 1 minus 0.63852.
And that's about 0.36148.
And three decimal places--
that would be 0.361.
If you look back in the upper right hand corner,
that's exactly what we got the other way.
And again, my contention is the 12.5 technique
for these problems is much easier than the 12.4 technique.