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The Practice of Statistics, 4th edition – For AP*

STARNES, YATES, MOORE

Chapter 5: Probability: What are the Chances?

Section 5.3

Conditional Probability and Independence

+ Chapter 5

Probability: What Are the Chances?

5.1 Randomness, Probability, and Simulation

5.2 Probability Rules

5.3 Conditional Probability and Independence

+ Section 5.3

Conditional Probability and Independence

After this section, you should be able to…

DEFINE conditional probability

COMPUTE conditional probabilities

DESCRIBE chance behavior with a tree diagram

DEFINE independent events

DETERMINE whether two events are independent

APPLY the general multiplication rule to solve probability questions

Learning Objectives

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What is Conditional Probability?

The probability we assign to an event can change if we know that some

other event has occurred. This idea is the key to many applications of

probability.

When we are trying to find the probability that one event will happen

under the condition that some other event is already known to have

occurred, we are trying to determine a conditional probability.

Definition:

The probability that one event happens given that another event

is already known to have happened is called a conditional

probability. Suppose we know that event A has happened.

Then the probability that event B happens given that event A

has happened is denoted by P(B | A).

Read | as “given that”

or “under the

condition that”

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Alternate Example: Who Owns a Home

High School Graduate?

Yes No Total

Homeowner 221 119 340

Not a Homeowner 89 71 160

Total 310 190 500

1. If we know that a person owns a home, what is the probability that the

person is a high school graduate? There are a total of 340 people in the

sample that own a home. Because there are 221 high school graduates

among the 340 home owners, the desired probability is

P(is a high school graduate given owns a home) = 221/340 or 65%

2. If we know that a person is a high school graduate, what is the probability

that the person owns a home? There are a total of 310 people who are

high school graduates. Because there are 221 home owners among the

310 high school graduates, the desired probability is

P(owns a home given is a high school graduate) = 221/310 or about 71%

+ Example: Grade Distributions

Consider the two-way table on page 314. Define events

E: the grade comes from an EPS course, and

L: the grade is lower than a B.

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Find P(L)

Find P(E | L)

Find P(L | E)

Total 3392 2952 3656 10000

Total

6300

1600

2100

P(L) = 3656 / 10000 = 0.3656

P(E | L) = 800 / 3656 = 0.2188

P(L| E) = 800 / 1600 = 0.5000

+ Conditional Probability and Independence

When knowledge that one event has happened does not change

the likelihood that another event will happen, we say the two

events are independent.

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Definition:

Two events A and B are independent if the occurrence of one

event has no effect on the chance that the other event will

happen. In other words, events A and B are independent if

P(A | B) = P(A) and P(B | A) = P(B).

P(left-handed | male) = 3/23 = 0.13

P(left-handed) = 7/50 = 0.14

These probabilities are not equal, therefore the

events “male” and “left-handed” are not independent.

Are the events “male” and “left-handed”

independent? Justify your answer.

Example:

+ Alternate Example: Allergies

Is there a relationship between gender and having allergies?

To find out, we used the random sampler at the United States

Census at School website (www.amstat.org/censusatschool)

to randomly select 40 US high school students who completed

a survey. The two-way table shows the gender of each

student and whether the student has allergies.

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P(allergies | female) = 10/23 = 0.435 P(allergies) = 18/40 = 0.45

These probabilities are close but not equal, therefore the events

“female” and “allergies” are not independent. Knowing that a student

was female slightly lowered the probability that she has allergies.

Are the events “female” and

“allergies” independent? Justify your

answer.

Female Male Total

Allergies 10 8 18

No Allergies 13 9 22

Total 23 17 40

+ Tree Diagrams

We learned how to describe the sample space S of a chance

process in Section 5.2. Another way to model chance

behavior that involves a sequence of outcomes is to construct

a tree diagram.

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Consider flipping a

coin twice.

What is the probability

of getting two heads?

Sample Space:

HH HT TH TT

So, P(two heads) = P(HH) = 1/4

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Picking Two Sneezers

In the previous alternate

example, we used a two-way

table that classified 40 students

according to their gender and

whether they had allergies.

Here is the table again.

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Suppose we chose 2 students

at random. What is the

probability that both students

suffer from allergies?

To get two students who suffer from allergies, we need to get an allergy sufferer

for the first student and an allergy sufferer for the second student. Following

along the top branches of the tree, we see that the probability is:

P(two allergy sufferers) = P(1st student has allergies and 2nd student has

allergies)

= P(1st student has allergies) P(2nd student has allergies | 1st student has

allergies)

= (18/40)(17/39) = 0.196

There is about a 20% chance of selecting two students with allergies.

Female Male Total

Allergies 10 8 18

No Allergies 13 9 22

Total 23 17 40

+ General Multiplication Rule

The idea of multiplying along the branches in a tree diagram

leads to a general method for finding the probability P(A ∩ B)

that two events happen together.

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The probability that events A and B both occur can be

found using the general multiplication rule

P(A ∩ B) = P(A) • P(B | A)

where P(B | A) is the conditional probability that event

B occurs given that event A has already occurred.

General Multiplication Rule

+ Example: Teens with Online Profiles

The Pew Internet and American Life Project finds that 93% of teenagers (ages

12 to 17) use the Internet, and that 55% of online teens have posted a profile

on a social-networking site.

What percent of teens are online and have posted a profile?

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P(online) 0.93

P(profile | online) 0.55

P(online and have profile) P(online) P(profile |online)

(0.93)(0.55)

0.5115

51.15% of teens are online and have

posted a profile.

+ Example: Playing in the NCAA

About 55% of high school students participate in a school athletic team at

some level and about 5% of these athletes go on to play on a college team in

the NCAA.

What percent of high school students play a sport in high school and go

on to play a sport in the NCAA?

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05.0)sport schoolhigh |sportNCAA (

55.0)sport schoolhigh (

P

P

)sport schoolhigh |sportNCAA ()sport schoolhigh (

)sportNCAA andsport schoolhigh (

PP

P

0275.0)05.0)(55.0(

Almost 3% of high school students will play a sport in high school

and in the NCAA.

+ Example: Who Visits YouTube?

See the example on page 320 regarding adult Internet users.

What percent of all adult Internet users visit video-sharing sites?

P(video yes ∩ 18 to 29) = 0.27 • 0.7

=0.1890

P(video yes ∩ 30 to 49) = 0.45 • 0.51

=0.2295

P(video yes ∩ 50 +) = 0.28 • 0.26

=0.0728

P(video yes) = 0.1890 + 0.2295 + 0.0728 = 0.4913

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Example: Media Usage and Good Grades

State: What percent of young people get good grades?

Plan: If we choose a subject in the

study at random, P(light user) =

0.17, P(moderate user) = 0.62,

P(heavy user) = 0.21, P(good

grades | light user) = 0.74, P(good

grades | moderate user) = 0.68,

P(good grades | heavy user) = 0.52.

We want to find the unconditional

probability P(good grades). A tree

diagram should help.

Do: There are three groups of students who get good grades, those who are light users and

get good grades, those who are moderate users and get good grades, and those who are

heavy users and get good grades. Because these groups are mutually exclusive, we can

add the probabilities of being in one of these three groups. P(good grades) = (0.17)(0.74) +

(0.62)(0.68) + (0.21)(0.52) = 0.1258 + 0.4216 + 0.1092 = 0.6566.

Conclude: About 66% of the students in the study described their grades as good.

In a study, 17% of the youth were classified as light media users, 62% were classified as

moderate media users and 21% were classified as heavy media users. Of the light users

who responded, 74% described their grades as good (A’s and B’s), while only 68% of the

moderate users and 52% of the heavy users described their grades as good. According to

this study, what percent of young people ages 8-18 described their grades as good?

+ Independence: A Special Multiplication Rule

When events A and B are independent, we can simplify the

general multiplication rule since P(B| A) = P(B).

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Definition:

Multiplication rule for independent events

If A and B are independent events, then the probability that A

and B both occur is

P(A ∩ B) = P(A) • P(B)

P(joint1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK)

=P(joint 1 OK) • P(joint 2 OK) • … • P(joint 6 OK)

=(0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87

Following the Space Shuttle Challenger disaster, it was determined that the failure

of O-ring joints in the shuttle’s booster rockets was to blame. Under cold

conditions, it was estimated that the probability that an individual O-ring joint would

function properly was 0.977. Assuming O-ring joints succeed or fail independently,

what is the probability all six would function properly?

Example:

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Example: First Trimester Screen

State: If 100 women with normal pregnancies are tested with the First Trimester

Screen, what is the probability that at least one woman will receive a positive

result?

Plan: It is reasonable to assume that the test results for different women are

independent. To find the probability of at least one false positive, we can use

the complement rule and the probability that none of the women will receive a

positive test result.

P(at least one positive) = 1 – P(no positive results)

Do: For women with normal pregnancies, the probability that a single test is not

positive is 1 – 0.05 = 0.95. The probability that all 100 women will get negative

results is (0.95)(0.95) (0.95) = (0.95)100 = 0.0059. Thus, P(at least one positive) =

1 – 0.0059 = 0.9941.

Conclude: There is over a 99% probability that at least one of the 100 women with

normal pregnancies will receive a false positive on the First Trimester Screen.

The First Trimester Screen is a non-invasive test given during the first trimester of

pregnancy to determine if there are specific chromosomal abnormalities in the fetus.

According to a study published in the New England Journal of Medicine in November 2005

(http://www.americanpregnancy.org/prenataltesting/firstscreen.html), approximately 5% of

normal pregnancies will receive a positive result. Among 100 women with normal

pregnancies, what is the probability that there will be at least one false positive?

+ Calculating Conditional Probabilities

If we rearrange the terms in the general multiplication rule, we

can get a formula for the conditional probability P(B | A).

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P(A ∩ B) = P(A) • P(B | A)

General Multiplication Rule

To find the conditional probability P(B | A), use the formula

Conditional Probability Formula

P(A ∩ B) P(B | A) P(A)

=

+ Example: Who Reads the Newspaper?

In Section 5.2, we noted that residents of a large apartment complex can be

classified based on the events A: reads USA Today and B: reads the New

York Times. The Venn Diagram below describes the residents.

What is the probability that a randomly selected resident who reads USA

Today also reads the New York Times?

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P(B | A) P(AB)

P(A)

P(AB) 0.05

P(A) 0.40

P(B | A) 0.05

0.40 0.125

There is a 12.5% chance that a randomly selected resident who reads USA

Today also reads the New York Times.

+ Example: Who Reads the Newspaper?

In an alternate example in section 5.2, we classified US households according

to the types of phones they used.

What is the probability that a randomly selected household with a landline

also has a cell phone?

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)(

) and ()|(

landlineP

landlinecellphonePlandlinecellphoneP

77.078.0

60.0

There is a 77% chance that a landline user also has a cell

phone.

Cell

Phone

No Cell

Phone Total

Landline 0.60 0.18 0.78

No

Landline 0.20 0.02 0.22

Total 0.80 0.20 1.00

We want to find P(cell phone | landline).

Using the conditional probability

formula,

+ Section 5.3

Conditional Probability and Independence

In this section, we learned that…

If one event has happened, the chance that another event will happen is a conditional probability. P(B|A) represents the probability that event B occurs given that event A has occurred.

Events A and B are independent if the chance that event B occurs is not affected by whether event A occurs. If two events are mutually exclusive (disjoint), they cannot be independent.

When chance behavior involves a sequence of outcomes, a tree diagram can be used to describe the sample space.

The general multiplication rule states that the probability of events A and B occurring together is P(A ∩ B)=P(A) • P(B|A)

In the special case of independent events, P(A ∩ B)=P(A) • P(B)

The conditional probability formula states P(B|A) = P(A ∩ B) / P(A)

Summary

+ Looking Ahead…

We’ll learn how to describe chance processes using the

concept of a random variable.

We’ll learn about

Discrete and Continuous Random Variables

Transforming and Combining Random Variables

Binomial and Geometric Random Variables

In the next Chapter…