Chapter 3 Conditional Probability and Independence Section 3

Chapter 3. Conditional Probability and Independence Section 3. 3. Theorem of Total Probability and Bayes’ Rule Section 3. 4 Odds, Odds Ratios, and Relative Risk Jiaping Wang Department of Mathematical Science 01/30/2013, Wednesday The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Outline Theorem of Total Probability Bayes’ Rule Odds, Odds Ratios and Relative Risk Homework #3 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Part 1. Theorem of Total Probability The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Theorem 3. 2 Consider an example, if there is a partition B 1 and B 2 such that B 1 UB 2=S and B 1∩B 2=ø, then we can find A = (A∩B 1) U (A∩B 2) and thus P(A)=P(A∩B 1) +P(A∩B 2)=P(A|B 1)P(B 1)+P(A|B 2)P(B 2) Theorem of Total Probability: If B 1, B 2, …, Bk is a collection of mutually exclusive and exhaustive events, then for any event A, we have The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Example 3. 8 A company buys microchips from three suppliers-I, II, and III. Supplier I has a record of providing microchips that contain 10% defectives; Supplier II has a defective rate of 5% and Supplier III has a defective rate of 2%. Suppose that 20%, 35% and 45% of the current supply came from Suppliers I, II, and III, respectively. If a microchip is selected at random from this supply, what is the probability that it is defective? Solution: BI={Chip comes from Supplier I}, BIII, D denote defective, ND – nondefective. P(BI∩D)=0. 20(0. 10)=0. 02, P(BI∩ND)=0. 18, P(BII∩D)=0. 175, P(BII∩ND)=0. 3325, P(BIII∩D)=0. 009, P(BIII∩ND)=0. 441. P(BI)=0. 20, P(BII)=0. 35, P(BIII)=0. 45. So by Law of Total Probability, P(D)=P(D|BI)P(BI)+P(D|BII)P(BII)+P(D|BIII)P(BIII)=0. 0465. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Part 2. Bayes’ Rule The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Theorem 3. 3 Bayes’ Rule. If the events B 1, B 2, …, Bk form a partition of the sample space S, and A is any event in S, then The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Example 3. 9 Consider again the information from Example 3. 8. If a random selected microchip is defective, what is the probability that it came from Supplier II? Solution: By Bayes’ rule, P(BII|D)=P(D|BII)P(BII)/P(D)=0. 376 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Part 3. Odds, Odds Ratios, and Relative Risk The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

An Example The Physicians’ Health study on the effects of aspirin on heart attacks randomly assigned to 22, 000 male physicians to either the “aspirin” or “placebo” arm of the study. The data on myocardial infarctions (MI) are given in table. MI No MI Total Aspirin 139 10, 898 11, 037 Placebo 239 10, 795 11034 Total 378 21, 683 22, 071 One my talk about what are the odds in favor of MI over non-MI. The odds in favor of an event A is the ratio of the probability of A to the probability of the complement of A. For aspirin P(MI)/P(non-MI)=(139/11037)/(10898/11037)=139/10898=0. 013; For placebo, P(MI)/P(non-MI)=239/10796=0. 022, which shows odds of heart attach with placebo is higher than the risk with aspirin, the odds ratio = Odds of MI with aspirin/odds of MI without aspirin=0. 59<1. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Definitions of Odds, Odds Ratios, and Relative Risk Odds ratios form a useful summary of the frequencies in a 2 x 2(two-way) frequency table. Yes No A a b B c d The odds in favor of A =a/b, odds in favor of B=c/d The odds ratio = a*d/b*c. The relative risk is the ratio of the probability of an event in the treatment group to the probability of same event in the placebo group, Relative risk = P(Yes|A)/P(Yes|B)=[a/(a+b)]/[c/(c+d)]=a(c+d)/c(a+b). The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Example 3. 10 The Physicaians’ Health Study included only men and the results clearly indicated that taking a low dose of aspirin reduced the risk of MI. In 2005, the results of the Women’s Health study were published in the table. This study randomized almost 40, 000 women, ages 45 and older, to either aspirin or placebo and followed the women for 10 years. MI No MI Total Aspirin 198 19, 736 19, 934 Placebo 193 19, 749 19, 942 Total 391 39, 485 39, 876 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Example 3. 10 Continue a. Find the odds of MI for the aspirin group. b. Find the odds of MI for the placebo group. c. Find the odds ratio of MI for the aspirin and placebo groups. d. Find the relative risk of MI for the aspirin and placebo group. Solutions: a. Odds for the aspirin: P(MI)/P(non-MI)=198/19736 b. Odds for the placebo: P(MI)/P(non-MI)=193/19749 c. Odds ratio = 198/19736*19749/193=1. 01 > 1 d. Relative risk = 198/19934*19942/193=1. 03>1 Which means low-dose aspirin regime is not effective for reducing MI for women. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Homework 3 Page 66: 3. 2 Page 67: 3. 6, 3. 11, 3. 14 Page 76: 3. 27 Page 77: 3. 32, 3. 34, 3. 36 Page 81: 3. 44 Page 86: 3. 55 Due on Wednesday, 02/06/2013. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL