1. Chapter 5: Probability: What are the Chances?
Section 5.3
Conditional Probability and Independence

2. 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”

3. Conditional Probability and Independence
Example: Grade Distributions
Consider the two-way table below. 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 =

4. Order and Conditional Probability
Order matters when asking conditional probability questions, just as it does with other mathematical operations such as subtraction or division.

5. 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.

6. Conditional Probability and Independence
Two events are independent if knowing the outcome of one event does not give you any additional information about the probability that the other event will occur.
This is similar to a scatterplot that has no association. When there is no association between the variables, knowing the value of the x variable won’t help you predict the value of the y variable.

7. Checkpoint
For each chance process below, determine whether the events are independent. Justify your answer.
1) Shuffle a standard deck of cards, and turn over the top card. Put it back in the deck, shuffle again, and turn over the top card. Define events A: first card is a heart, and B: second card is a heart.
2) Shuffle a standard deck of cards, and turn over the top two cards, one at a time. Define events A: first card is a heart, and B: second card is a heart.
Independent. Since we are putting the 1st card back and then reshuffling the cards before drawing the 2nd card, knowing what the 1st card was will not tell us anything about what the 2nd card will be.
Not independent. Once we know the suit of the 1st card, then the probability of getting a heart on the 2nd card will change depending on what the 1st card was.

8. Checkpoint
For each chance process below, determine whether the events are independent. Justify your answer.
3) The 28 students in an AP Statistics class completed a brief survey. One of the questions asked whether each student was right- or left- handed. The two way table summarizes the class data. Choose a student from the class at random. The events of interest are “female” and “right-handed.”
Handedness
Female
Male
Left 3 1
Right 18 6
Independent. Once we know that the chosen person is female, this does not tell us anything more about whether she is right-handed or not. Overall, 24/28 = 6/7 of the students are right-handed. And, among the women, 18/21 = 6/7 are right-handed.
So, P(right-handed) = P(right-handed │ female).

9. Independence Are the following events independent?
1) A: heads on a coin
B: roll a 4 on a fair die
2) A: choosing a heart from a deck of cards
B: choosing a 5 from a deck of cards
Done with replacement
3) #2 done without replacement
4) Two students in our class are chosen for an all expenses paid trip to 10th period.
A: student has blond hair
B: student is a girl

10. 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

11. Example – Tree Diagrams in Action
The two-way table showing the gender and handedness of the students in the class is repeated below: Problem: Suppose we choose two students at random A) Draw a tree diagram that shows the sample space for this chance process.
Handedness
Female
Male
Left 3 1
Right 18 6

12. Example – Tree Diagrams in Action
The two-way table showing the gender and handedness of the students in the class is repeated below: Problem: Suppose we choose two students at random B) Find the probability that both students are left-handed.
Handedness
Female
Male
Left 3 1
Right 18 6
To get two left-handed students, we’d have to get a lefty for the first student chosen and a lefty for the second student chosen. Following the top branches of the tree, we see that the probability is
P(two lefties) = P(1st student lefty and 2nd student is left)
= P(1st student lefty) ● P(2nd student lefty │ 1st student lefty)
=4/28 ● 3/27 ≈ 0.016
There’s about a 1.6% chance of randomly selecting two lefties.

13. 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

14. 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.

15. Example: Who Visits YouTube?
Video-sharing sites, led by YouTube, are popular destinations on the Internet. Let’s look only at adult Internet users, aged 18 and over. About 27% of adult Internet users are 18 to 29 years old, another 45% are 30 to 49 years old, and the remaining 28% are 50 and over. The Pew Internet and American Life Projects finds that 70% of Internet users aged 18 to 29 have visited a video-sharing site, along with 51% of those aged 30 to 49 and 26% of those 50 or older. Do most Internet users visit YouTube and similar sites?

16. Example: Who Visits YouTube?
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) = =

17. Checkpoint 1) Construct a tree diagram to represent this situation.
A computer company makes desktop and laptop computers at factories in three states – California, Texas, and New York. The California factory produces 40% of the company’s computers, the Texas factory makes 25%, and the remaining 35% are manufactured in New York. Of the computers made in California, 75% are laptops. Of those made in Texas and New York, 70% and 50%, respectively, are laptops. All computers are first shipped to a distribution center in Missouri before being sent out to stores. Suppose we select a computer at random from the distribution center.
1) Construct a tree diagram to represent this situation.

18. Checkpoint
2) Find the probability that the computer is a laptop. Show your work.
P(laptop) = P(laptop ∩ CA) + P(laptop ∩ TX) + P(laptop ∩ NY)
= 0.65

19. 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

20. Independence and Mutually Exclusive
Select one card from a standard deck and define the events A: the card is red and B: the card is a club.
Since there are no red clubs, A and B are mutually exclusive. However, since P(A) = 0.5 and P(A │ B) = 0, these events are not independent.
Select one card from a standard deck and define the events A: the card is red and B: the card is a 7.
Since there are two red 7s, A and B are not mutually exclusive. However, since P(A) = 0.5 and P(A │ B) = 0.5, these events are independent.
Select one card from a standard deck and define the events A: the card is red and B: the card is a heart.
Since all the hearts are red, A and B are not mutually exclusive. Also, since P(A) = 0.5 and P(A │ B) = 1, these events are not independent.

21. Checkpoint
1) During World War II, the British found that the probability that a bomber is lost through enemy action on a mission over occupied Europe was Assuming that missions are independent, find the probability that a bomber returned safely from 20 missions.
P(one returned safely) = 0.95
Assuming independence, P(safe return on all 20 missions) = (0.95)20 =

22. Checkpoint
2) Government data show that 8% of adults are full- time college students and that 30% of adults are age 55 or older. Since (0.08)(0.30) = 0.024, can we conclude that about 2.4% of adults are college students 55 or older? Why or why not?
No. Whether one is a college student and one’s age are not independent events. Far more younger people than older people are college students.

23. 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

24. 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(A | B).
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(A | B), use the formula
Conditional Probability Formula

25. Conditional Probability and Independence
Example: Who Reads the Newspaper?
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.

26. Example – Conditioning “backward” in a Tree Diagram
In an earlier example, we looked at whether adult Internet users of various ages regularly visit video-sharing sites like YouTube. The tree diagram summarizing the probabilities is shown below
What percent of adult internet users who visit video-sharing sites are aged 18 to 29?
Follow the 4-step process.

27. State: What percent of adult Internet users who visit
video-sharing sites are aged 18 to 29?
Plan: In probability language, we want the conditional
probability P(18 to 29 │ video yes). Use the tree
diagram and the definition of conditional
probability:

28. Do: Look at the tree diagram. P(video yes) is the sum
of the three red probabilities. P(18 to 29 and
video yes) is the result of following just the top
branch in the tree diagram. So
Conclude: About 38% of adults who visit video-sharing sites are between 18 and 29 years old. Compare this conditional probability with the original information (unconditional) that 27% of adult internet users are between 18 and 29 years old. Knowing that a person visits video-sharing sites increases the probability that he or she is young (which indicates that these two events are not independent.

29. Example – Conditional Probability in Real Life
Many employers require prospective employees to take a drug test. A positive result on this test indicates that the prospective employee uses illegal drugs. However, not all people who test positive actually use drugs. Suppose that 4% of prospective employees use drugs, the false positive rate is 5%, and the false negative rate is 10%.
What is the percent of people who test positive actually use illegal drugs?

30. Solution: A tree diagram best summarizes this situation.

31. P(took drugs │ positive test) =
Two groups test positive: (1) those who take drugs and test positive (probability = 0.036) and (2) those who don’t take drugs and test positive (probability = 0.048), so the probability of testing positive is = Thus,
That is, 42.9% of prospective employees who test positive actually took drugs.

32. Common Test Error
Many students lose credit on probability questions because they do not show their work. When using the conditional probability formula, show the numerator and the denominator of the probability, not just the decimal answer.

33. Solving Conditional Probability Problems
Most conditional probability questions can be solved using a tree diagram or a two-table.
If the problem provides conditional probabilities, it is often easier to use a tree diagram.
If the problem provides counts or proportions of people in different catergories, a two-way table is often easier to use.

34. 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)