1. Chapter 5: Probability: What are the Chances?
Section 5.3
Conditional Probability and Independence
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE

2. Chapter 5 Probability: What Are the Chances?
5.1 Randomness, Probability, and Simulation
5.2 Probability Rules
5.3 Conditional Probability and Independence

3. Section 5.3 Conditional Probability and Independence
Learning Objectives
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

4. Conditional Probability and Independence
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.
Conditional Probability and Independence
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”

5. Conditional Probability and Independence
Alternate Example: Who Owns a Home
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%
High School Graduate?
Yes No
Total
Homeowner
221
119
340
Not a Homeowner 89 71
160
310
190
500
Conditional Probability and Independence
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%

6. Conditional Probability and Independence
Example: Grade Distributions
Consider the two-way table on page Define events
E: the grade comes from an EPS course, and
L: the grade is lower than a B.
Conditional Probability and Independence
Total
Total
6300
1600
2100
Find P(L)
Find P(E | L)
Find P(L | E)
P(L) = 3656 / =
P(E | L) = 800 / 3656 =
P(L| E) = 800 / 1600 =

7. 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.
Conditional Probability and Independence
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).
Are the events “male” and “left-handed” independent? Justify your answer.
Example:
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.

8. Conditional Probability and Independence
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 ( 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.
Conditional Probability and Independence
Are the events “female” and “allergies” independent? Justify your answer.
Female
Male
Total
Allergies 10 8 18
No Allergies 13 9 22 23 17 40
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.

9. Conditional Probability and Independence
Tree Diagrams
We learned how to describe the sample space S of a chance process in Section Another way to model chance behavior that involves a sequence of outcomes is to construct a tree diagram.
Conditional Probability and Independence
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

10. Conditional Probability and Independence
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.
Female
Male
Total
Allergies 10 8 18
No Allergies 13 9 22 23 17 40
Conditional Probability and Independence
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.

11. Conditional Probability and Independence
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.
Conditional Probability and Independence
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

12. Conditional Probability and Independence
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?
Conditional Probability and Independence
51.15% of teens are online and have posted a profile.

13. Conditional Probability and Independence
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?
Conditional Probability and Independence
Almost 3% of high school students will play a sport in high school and in the NCAA.

14. 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) = • 0.7
=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) = =

15. Example: Media Usage and Good Grades
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?
State: What percent of young people get good grades?
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) = =
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) = We want to find the unconditional probability P(good grades). A tree diagram should help.
Conclude: About 66% of the students in the study described their grades as good.

16. Conditional Probability and Independence
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).
Conditional Probability and Independence
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)
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 Assuming O-ring joints succeed or fail independently, what is the probability all six would function properly?
Example:
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

17. Example: 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 ( 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?
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 = The probability that all 100 women will get negative results is (0.95)(0.95) (0.95) = (0.95)100 = Thus, P(at least one positive) = 1 – =
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.

18. Conditional Probability and Independence
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).
Conditional Probability and Independence
General Multiplication Rule
P(A ∩ B) = P(A) • P(B | A)
P(A ∩ B)
P(A)
P(B | A)
To find the conditional probability P(B | A), use the formula
Conditional Probability Formula =

19. Conditional Probability and Independence
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?
Conditional Probability and Independence
There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.

20. Conditional Probability and Independence
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?
We want to find P(cell phone | landline). Using the conditional probability formula,
Conditional Probability and Independence
Cell Phone
No Cell Phone
Total
Landline
0.60
0.18
0.78
No Landline
0.20
0.02
0.22
0.80
1.00
There is a 77% chance that a landline user also has a cell phone.

21. Section 5.3 Conditional Probability and Independence
Summary
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)

22. Looking Ahead… In the next Chapter…
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…