The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Chapter 3. Conditional Probability and Independence Section 3.3. Theorem of Total Probability and Bayes’

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The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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 B1 and B2 such that B1UB2=S and B1∩B2=ø, then we can find A = (A∩B1) U (A∩B2) and thus P(A)=P(A∩B1) +P(A∩B2)=P(A|B1)P(B1)+P(A|B2)P(B2) Theorem of Total Probability: If B 1, B 2, …, B k is a collection of mutually exclusive and exhaustive events, then for any event A, we have Theorem of Total Probability: If B 1, B 2, …, B k is a collection of mutually exclusive and exhaustive events, then for any event A, we have

The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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? 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? Example 3.8 Solution: BI={Chip comes from Supplier I}, BII, BIII, D denote defective, ND – non- defective. 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)= 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)=

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 B1, B2, …, Bk form a partition of the sample space S, and A is any event in S, then Bayes’ Rule. If the events B1, B2, …, 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? 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. 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. MINo MITotal Aspirin13910,89811,037 Placebo23910, Total37821,68322,071 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 YesNo Aab Bcd Odds ratios form a useful summary of the frequencies in a 2x2(two-way) frequency table. 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. MINo MITotal Aspirin19819,73619,934 Placebo19319,74919,942 Total39139,48539,876

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. 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. 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.