MATH 110 Sec 12-4 (S2020): Addition and Complement Rules

Added some problems to the end of the Fall 2019 video

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.