1. CHAPTER 12 General Rules of Probability BPS - 5TH ED.CHAPTER 12 1

2. PROBABILITY RULES FROM CHAPTER 10 BPS - 5TH ED.CHAPTER 12 2

3. VENN DIAGRAMS BPS - 5TH ED.CHAPTER 12 3 Two disjoint events: Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common:

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 4

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: ? BPS - 5TH ED.CHAPTER 12 5 } 68% 32% (100%  68%) What if pairs are self-selected?

6. ADDITION RULE: FOR DISJOINT EVENTS BPS - 5TH ED.CHAPTER 12 6 P(A or B) = P(A) + P(B)

7. GENERAL ADDITION RULE BPS - 5TH ED.CHAPTER 12 7 P(A or B) = P(A) + P(B)  P(A and B)

8. CASE STUDY BPS - 5TH ED.CHAPTER 12 8 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?

9. CASE STUDY BPS - 5TH ED.CHAPTER 12 9 Other Students P(A or B)= P(A) + P(B)  P(A and B) = 0.80 + 0.30  0.20 = 0.90 All Students Part-time (B) 0.30 {A and B} 0.20 In-state (A) 0.80

10. CASE STUDY BPS - 5TH ED.CHAPTER 12 10 Other Students All Students Part-time (B) 0.30 {A and B} 0.20 In-state (A) 0.80 In-state, but not part-time (A but not B): 0.80  0.20 = 0.60

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 11

12. CONDITIONAL PROBABILITY When P(A) > 0, the conditional probability of B given A is BPS - 5TH ED.CHAPTER 12 12

13. CASE STUDY BPS - 5TH ED.CHAPTER 12 13 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)?

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: BPS - 5TH ED.CHAPTER 12 14 P(A and B) = P(A)  P(B|A) or P(A and B) = P(B)  P(A|B)

15. CASE STUDY BPS - 5TH ED.CHAPTER 12 15 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?

16. CASE STUDY BPS - 5TH ED.CHAPTER 12 16 Student Demographics P(A) = 0.25, P(B|A) = 0.20 5% of all students are freshmen smokers. P(A and B)= P(A)  P(B|A) = 0.25  0.20 = 0.05

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 17