Griffiths D. Head First Statistics

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calculating probabilities

P(A ∩ B) + P(A ∩ B

)

P(A

∩ B

)

P(A ∪ B) - P(B)

BE the probability

Your job is to play like you’re the

probability and shade in the area

that represents each of the following

probabilities on the Venn

diagrams.

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A B

P(A ∩ B) + P(A ∩ B

)

P(A

∩ B

)

P(A ∪ B) - P(B)

BE the probability Solution

Your job was to play like you’re the probability

and shade in the area that represents each of

the probabilities on the Venn diagrams.

be the probability solution

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calculating probabilities

50 sports enthusiasts at the Head First Health Club are asked whether they play baseball,

football, or basketball. 10 only play baseball. 12 only play football. 18 only play basketball. 6 play

baseball and basketball but not football. 4 play football and basketball but not baseball.

Draw a Venn diagram for this probability space. How many enthusiasts play baseball in total?

How many play basketball? How many play football?

Are any sports’ rosters mutually exclusive? Which sports are exhaustive (fill up the possibility

space)?

A or B

To find the probability of getting event

A or B, use

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

∪ means OR

∩ means AND

Vital StatisticsVital Statistics

154 Chapter 4

50 sports enthusiasts at the Head First Health Club are asked whether they play baseball, football

or basketball. 10 only play baseball. 12 only play football. 18 only play basketball. 6 play baseball

and basketball but not football. 4 play football and basketball but not baseball.

Draw a Venn diagram for this probability space. How many enthusiasts play baseball in total? How

many play basketball? How many play football?

Are any sports’ rosters mutually exclusive? Which sports are exhaustive (fill up the possibility

space)?

This information

looks complicated,

but drawing a

Venn diagram

will help us to

visualize what’s

going on.

By adding up the values in each circle in the Venn diagram, we can determine that there

are 16 total baseball players, 28 total basketball players, and 16 total football players.

The baseball and football events are mutually exclusive. Nobody plays both baseball and

football, so P(Baseball ∩ Football) = 0

The events for baseball, football, and basketball are exhaustive. Together, they fill the

entire possibility space, so P(Baseball ∪ Football ∪ Basketball) = 1

Are A and A

mutually exclusive or

exhaustive?

A: Actually they’re both. A and A

can

have no common elements, so they are

mutually exclusive. Together, they make

up the entire possibility space so they’re

exhaustive too.

Isn’t P(A ∩ B) + P(A ∩ B

) just a

complicated way of saying P(A)?

A: Yes it is. It can sometimes be useful

to think of different ways of forming the

same probability, though. You don’t always

have access to all the information you’d

like, so being able to think laterally about

probabilities is a definite advantage.

Is there a limit on how many events

can intersect?

A: No. When you’re referring to the

intersection between several events, use

∩‘s. As an example, the intersection of

events A, B, and C is A

∩ B ∩ C.

Finding probabilities for multiple

intersections can sometimes be tricky. We

suggest that if you’re in doubt, draw a Venn

diagram and take a good, hard look at which

probabilities need to be added together and

which need to be subtracted.

Baseball

Football

Basketball

The numbers we’ve

been given all add

up to 50, the

total number of

sports lovers.

exercise solution

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calculating probabilities

…but it’s time for another bet

Even with the odds in our favor, we’ve been unlucky with the outcomes at

the roulette table. The croupier decides to take pity on us and offers a little

inside information. After she spins the roulette wheel, she’ll give us a clue

about where the ball landed, and we’ll work out the probability based on

what she tells us.

Here’s your next

bet…and a hint about

where the ball landed.

Shh, don’t tell Fat Dan...

Bet: Even

Clue: The ball

landed in a

black pocket

Another unlucky spin…

We know that the probability of the ball landing on black or even

is 0.684, but, unfortunately, the ball landed on 23, which is red and

odd.

Should we take this bet?

How does the probability of getting even given that

we know the ball landed in a black pocket compare

to our last bet that the ball would land on black or

even. Let’s figure it out.

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108 10

108 810

Conditions apply

The croupier says the ball has landed in a black pocket.

What’s the probability that the pocket is also even?

Black Even

We already know

the pocket is black.

We want the probability

that the pocket is even,

given that it’s black.

But we’ve

already done this;

it’s just the probability of

getting black and even.

This is a slightly different problem

We don’t want to find the probability of getting a pocket

that is both black and even, out of all possible pockets.

Instead, we want to find the probability that the pocket is

even, given that we already know it’s black.

We can ignore these

areas—we know that

the pocket is black.

In other words, we want to find out how many pockets

are even out of all the black ones. Out of the 18 black

pockets, 10 of them are even, so

Black

Even

P(Even giv

en Black) = 10

= 0.556 (to 3 decimal places)

10 out of

18 are even.

introducing conditional probability

It turns out that even with some inside information, our odds are

actually lower than before. The probability of even given black is

actually less than the probability of black or even.

However, a probability of 0.556 is still better than 50% odds, so

this is still a pretty good bet. Let’s go for it.

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calculating probabilities

P(A | B) = P(A ∩ B)

P(B)

B A

I’m a given

P(A ∩ B)

P(B)

We can rewrite this equation to give us a way of finding P(A ∩ B)

P(A ∩ B) = P(A | B)

P(B)

It doesn’t end there. Another way of writing P(A ∩ B) is P(B ∩ A).

This means that we can rewrite the formula as

P(B ∩ A) = P(B | A)

P(A)

In other words, just flip around the A and the B.

Find conditional probabilities

So how can we generalize this sort of problem? First of

all, we need some more notation to represent conditional

probabilities, which measure the probability of one event

occurring relative to another occurring.

If we want to express the probability of one event happening

given another one has already happened, we use the “|” symbol

to mean “given.” Instead of saying “the probability of event A

occurring given event B,” we can shorten it to say

P(A | B)

So now we need a general way of calculating P(A | B). What

we’re interested in is the number of outcomes where both A and

B occur, divided by all the B outcomes. Looking at the Venn

diagram, we get:

It looks like it can be difficult to show

conditional probability on a Venn diagram.

I wonder if there’s some other way.

Venn diagrams aren’t always the best way of

visualizing conditional probability.

Don’t worry, there’s another sort of diagram you can use—a

probability tree.

Because we’re trying to find the

probability of A given B, we’re only

interested in the set of events

where B occurs.

The probability of A given

that we know B has happened.

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It’s not always easy to visualize conditional probabilities with

Venn diagrams, but there’s another sort of diagram that really

comes in handy in this situation—the probability tree.

Here’s a probability tree for our problem with the roulette

wheel, showing the probabilities for getting different colored

and odd or even pockets.

Black

Red

Green

Even

Odd

Even

Odd

18/38

2/38

8/18

10/18

1/2

These are branches,

like the branches

of a tree.

Here are the first set of

exclusive events, the colors. The

probabilities for each event go

along the relevant branch.

These are all the

second set of events.

The probabilities for

each set of branches

must add up to 1.

P(Green)

P(00 | Green)

The first set of branches shows the probability of each

outcome, so the probability of getting a black is 18/38, or

0.474. The second set of branches shows the probability

of outcomes given the outcome of the branch it is

linked to. The probability of getting an odd pocket given

we know it’s black is 8/18, or 0.444.

probability trees

You can visualize conditional probabilities with a probability tree

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calculating probabilities

P(B)

P(A | B)

Probability trees don’t just help you visualize probabilities; they can help

you to calculate them, too.

Let’s take a general look at how you can do this. Here’s another

probability tree, this time with a different number of branches. It shows

two levels of events: A and A

and B and B

. A

refers to every possibility

not covered by A, and B

refers to every possibility not covered by B.

You can find probabilities involving intersections by multiplying the

probabilities of linked branches together. As an example, suppose you

want to find P(A ∩ B). You can find this by multiplying P(B) and P(A | B)

together. In other words, you multiply the probability on the first level B

branch with the probability on the second level A branch.

P(A ∩ B) = P(A | B)

P(B)

P(A

| B)

P(A

∩ B) = P(A

| B)

P(B)

P(A | B

)

P(A ∩ B

) = P(A | B

)

P(B

)

P(A

| B

)

P(A

∩ B

) = P(A

| B

)

P(B

)

P(B

)

To find P(A

∩

B), just multiply

the probabilities for these two

branches together.

The probability of not

getting A given that B

hasn’t happened

This is the same equation you

saw earlier—just multiply the

adjoining branches together.

The probability of

not getting event B

Using probability trees gives you the same results you saw earlier, and

it’s up to you whether you use them or not. Probability trees can be time-

consuming to draw, but they offer you a way of visualizing conditional

probabilities.

Trees also help you calculate conditional probabilities

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Donuts

Coffee

Donuts

Coffee

Probability Magnets

Duncan’s Donuts are looking into the probabilities of their customers buying

donuts and coffee. They drew up a probability tree to show the probabilities,

but in a sudden gust of wind, they all fell off. Your task is to pin the

probabilities back on the tree. Here are some clues to help you.

3/4

1/4

2/5

3/5

1/3

2/3

P(Donuts) = 3/4 P(Coffee | Donuts

) = 1/3 P(Donuts ∩ Coffee) = 9/20

probability magnets