1. Chapter 6: Probability: What are the Chances?
Section 6.3
General Probability Rules
The Practice of Statistics, 3rd edition – For AP*
STARNES, YATES, MOORE

2. Section 6.3 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

3. Addition Rule for Disjoint Events
If two events A and B are disjoint:
P(A or B) = P(A) + P(B)
If three events are disjoint:
P(A or B or C) = P(A) + P(B) + P(C)
P(A or B) = P (A) + P(B)
Can also be written P(A B).
This is the union of events A and B. A B

4. Mutually Exclusive (Disjoint) Events
Two events have no outcomes in common.
Example: An animal can’t be a dog and a cat at the same time.
Example: Roll a “2” or a “5” – these can’t happen at the same time.
Note: the second “or” statement involves only one roll of the die, not two rolls.

5. Non –Disjoint Events: Can Have Outcomes in Common.
If two events E & F are not disjoint,
P(E or F) = P(E) + P(F) – P(E and F)
This is The General Addition Rule.
P(E and F) is called a joint probability and can be written P(E∩F).
P(E∩F) is also called the intersection of events E and F.

6. Non Disjoint Events
The shaded region shows the intersection of events A and B, where A and B can happen at the same time. A B
EXAMPLE:
Event A = (Being a senior at KLHS)
Event B = (Taking Statistics)
You can have seniors at KLHS who are taking Stats! These events can happen at the same time.

7. Example 6.23 on page 438
P (Deb becoming a partner) = .7 P (Matt becoming a partner) = .5 P (Deb and Matt becoming a partner) = .3 P(Deb or Matt becoming a partner) = 1.2?? This probability exceeds 1 which is impossible!

8. Venn Diagram Interpretation
D and M
D and MC
DC and M
DC and MC
Let’s add in all the values for these 4 joint probabilities!

9. Example 6.17 Continued
P(at least one is promoted) = P(Deb or Matt) = = .9 P(neither is promoted) = 1-.9 = .1 The reason that we “correct” by subtracting .3 (the intersection of Deb and Matt) is that if we don’t, it is counted twice.

10. Union
The union of event A with event B consists of all outcomes that are in at least one of the two events.
Example: Event E is rolling a die and getting prime number or even number.
E = {2,3,4,5,6}

11. Venn Diagram - A or B P(A) consists of {2,3,5}.
P(B) consists of {2,4,6}.
A U B consists of {2,3,4,5,6}

12. Intersection This is the event where both A and B happen.
It consists of all outcomes that are in both events.
Denoted:

13. Venn Diagram - A and B A B
From the previous example, A∩B is {2}, which is both
an even and prime number.

14. General Rule for the Union of Two Events
P(A or B) = P(A) + P(B) – P(A and B).
Note: if there is no intersection (if A and B are mutually exclusive), then the term:
P(A and B) is equal to zero, which returns us to the Addition Rule for Disjoint Events.

15. Homework # Pg 440 #65, 67-69, 70 General Probability Rules Part 1

19. Chapter 6: Probability: What are the Chances?
In an apartment complex, 40% of residents read USA Today. Only 25% read the New York Times. Five percent of residents read both papers. Suppose we select a resident of the apartment complex at random and record which of the two papers the person reads.
Construct a Venn diagram to represent the outcomes of this chance process.
Find the probability that the person reads at least one of the two papers.
Find the probability that the person doesn’t read either paper.
(b) If the randomly selected person reads at least one of the two papers, then he or she reads USA Today, the New York Times, or both papers. But that’s the same as the event A ∪ B. From the two-way table, the Venn diagram, or the general addition rule, we have P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = − 0.05 = 0.60 So there’s a 60% chance that the randomly selected resident reads at least one of the two papers.
(c) From the two-way table or Venn diagram, P(reads neither paper) = P(AC ∩ BC ) = 0.40.
Chapter 6: Probability: What are the Chances?
Section 6.3
Conditional Probability and Independence
The Practice of Statistics, 3rd edition – For AP*
STARNES, YATES, MOORE

20. 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”

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

22. Conditional Probability and Independence
Example: You Try!
Conditional Probability and Independence
19/90
4/88
84/103

23. Conditional Probability and Independence
General Multiplication Rule
Be sure you understand how we found the three probabilities. There is a relationship among these probabilities.
P(A and B) = P(A) X P(B|A)
First the grade from EPS; then, given that it is from EPS, it is below B. We have discovered the 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

24. 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

25. 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.

26. Are these events independent?
Junior and AP Calc?
Senior and AP Stats?
Earns A in AP Stats
Earns A in AP Calc
Total
Junior 5 7 12
Senior 9 21 17 16 33
P (Senior│𝐴𝑃 Stats) = 12/17 ; P(Senior)= 21/33
Since the values are not equal, the events are not independent.

27. Conditional Probability and Independence
Example: Who Reads the Newspaper?
In Section 6.2, 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? P(A ∩ B) = P(A) • P(B | A)
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.

28. 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 =

29. HW Pg 446 #71-73, 76, 77 General Probability Rules Part 1

32. Chapter 6: Probability: What are the Chances?
Slim is playing poker and holds 3 diamonds. He hopes to draw two diamonds in a row in order to get a flush (all cards of the same suit). Of all the cards showing (those in Slim’s hand and those upturned on the table), he sees 11 cards. 4 of those 11 cards are diamonds. So 9 of the 41 remaining unseen cards must be diamonds. What is the probability that Slim will draw his two diamonds?
P(first card is a diamond) = 9/41
P(second card diamond | given first card diamond) = 8/40
9/41 * 8/40 = 0.044
Chapter 6: Probability: What are the Chances?
Section 6.3
Conditional Probability and Independence
The Practice of Statistics, 3rd edition – For AP*
STARNES, YATES, MOORE

33. 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

34. Conditional Probability and Independence
REMEMBER: 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).

35. 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

36. 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.

37. Example: Who Visits YouTube?
Example 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) = =

38. Section 6.3 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)

39. HW Pg 451 #80-82, 87, 93 READ Pg 447 -451 Look at Examples
General Probability Rules Part 1
General Probability Rules Part 2