1. Basic Practice of Statistics - 3rd Edition
Chapter 12
General Rules of Probability
BPS - 5th Ed.
Chapter 12
Chapter 11

2. Probability Rules from Chapter 10
BPS - 5th Ed.
Chapter 12

3. Venn Diagrams Two disjoint events: Chapter 12
Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common:
Two disjoint events:
BPS - 5th Ed.
Chapter 12

4. Multiplication Rule for Independent Events
If two events A and B do not influence each other, and if knowledge about one does not change the probability of the other, the events are said to be independent of each other.
If two events are independent, the probability that they both happen is found by multiplying their individual probabilities:
P(A and B) = P(A)  P(B)
BPS - 5th Ed.
Chapter 12

5. Multiplication Rule for Independent Events Example
Suppose that about 20% of incoming male freshmen smoke.
Suppose these freshmen are randomly assigned in pairs to dorm rooms (assignments are independent).
The probability of a match (both smokers or both non-smokers):
both are smokers: 0.04 = (0.20)(0.20)
neither is a smoker: 0.64 = (0.80)(0.80)
only one is a smoker: ?
} 68%
32% (100%  68%)
What if pairs are self-selected?
BPS - 5th Ed.
Chapter 12

6. Addition Rule: for Disjoint Events
P(A or B) = P(A) + P(B)
BPS - 5th Ed.
Chapter 12

7. P(A or B) = P(A) + P(B)  P(A and B)
General Addition Rule
P(A or B) = P(A) + P(B)  P(A and B)
BPS - 5th Ed.
Chapter 12

8. Case Study Student Demographics
At a certain university, 80% of the students were in-state students (event A), 30% of the students were part-time students (event B), and 20% of the students were both in-state and part-time students (event {A and B}). So we have that P(A) = 0.80, P(B) = 0.30, and P(A and B) = 0.20.
What is the probability that a student is either an in-state student or a part-time student?
BPS - 5th Ed.
Chapter 12

9. Case Study 0.30 0.20 0.80 All Students
Other Students
Part-time (B)
0.30
In-state (A)
0.80
{A and B}
0.20
P(A or B) = P(A) + P(B)  P(A and B)
=  = 0.90
BPS - 5th Ed.
Chapter 12

10. In-state, but not part-time (A but not B): 0.80  0.20 = 0.60
Case Study
All Students
Other Students
Part-time (B)
0.30
In-state (A)
0.80
In-state, but not part-time (A but not B): 0.80  0.20 = 0.60
{A and B}
0.20
BPS - 5th Ed.
Chapter 12

11. Conditional Probability
The probability of one event occurring, given that another event has occurred is called a conditional probability.
The conditional probability of B given A is denoted by P(B|A)
the proportion of all occurrences of A for which B also occurs
BPS - 5th Ed.
Chapter 12

12. Conditional Probability
When P(A) > 0, the conditional probability of B given A is
BPS - 5th Ed.
Chapter 12

13. Case Study Student Demographics In-state (event A): P(A) = 0.80
Part-time (event B): P(B) = 0.30
Both in-state and part-time: P(A and B) = 0.20.
Given that a student is in-state (A), what is the probability that the student is part-time (B)?
BPS - 5th Ed.
Chapter 12

14. General Multiplication Rule
For ANY two events, the probability that they both happen is found by multiplying the probability of one of the events by the conditional probability of the remaining event given that the other occurs:
P(A and B) = P(A)  P(B|A)
or P(A and B) = P(B)  P(A|B)
BPS - 5th Ed.
Chapter 12

15. Case Study Student Demographics
At a certain university, 20% of freshmen smoke, and 25% of all students are freshmen.
Let A be the event that a student is a freshman, and let B be the event that a student smokes.
So we have that P(A) = 0.25, and P(B|A) = 0.20.
What is the probability that a student smokes and is a freshman?
BPS - 5th Ed.
Chapter 12

16. 5% of all students are freshmen smokers.
Case Study
Student Demographics
P(A) = 0.25 , P(B|A) = 0.20
P(A and B) = P(A)  P(B|A)
= 0.25  0.20
= 0.05
5% of all students are freshmen smokers.
BPS - 5th Ed.
Chapter 12

17. Independent Events
Two events A and B that both have positive probability are independent if P(B|A) = P(B)
General Multiplication Rule:
P(A and B) = P(A)  P(B|A)
Multiplication Rule for independent events:
P(A and B) = P(A)  P(B)
BPS - 5th Ed.
Chapter 12

18. Tree Diagrams
Useful for solving probability problems that involve several stages
Often combine several of the basic probability rules to solve a more complex problem
probability of reaching the end of any complete “branch” is the product of the probabilities on the segments of the branch (multiplication rule)
probability of an event is found by adding the probabilities of all branches that are part of the event (addition rule)
BPS - 5th Ed.
Chapter 12

19. Binge Drinking and Accidents
Case Study
Binge Drinking and Accidents
At a certain college, 30% of the students engage in binge drinking. Among college-aged binge drinkers, 18% have been involved in an alcohol-related automobile accident, while only 9% of non-binge drinkers of the same age have been involved in such accidents.
Let event A = {accident related to alcohol}.
Let event B = {binge drinker}.
So we have P(A|B)=0.18, P(A|’not B’)=0.09, & P(B)=0.30 .
What is the probability that a randomly selected student has been involved in an alcohol-related automobile accident?
BPS - 5th Ed.
Chapter 12

20. Binge Drinking and Accidents
Case Study
Binge Drinking and Accidents
P(A and B)
= P(B)P(A|B)
= (0.30)(0.18)
Accident
No accident
0.09
0.82
0.18
0.91
0.054
Binge drinker
Non-binge drinker
0.30
0.70
0.637
0.063
0.246
P(Accident) = P(A) = = 0.117
BPS - 5th Ed.
Chapter 12