1. Conditional Probability Lecture 33 Section 7.4.2 Tue, Mar 23, 2004

2. Conditional Probability Suppose that for a randomly selected day, Suppose that for a randomly selected day, P(wind) = 0.3. P(wind) = 0.3. P(rain) = 0.2. P(rain) = 0.2. P(wind and rain) = 0.1. P(wind and rain) = 0.1. Find P(wind | rain) and P(rain | wind). Find P(wind | rain) and P(rain | wind). Is it easier to find the answer Is it easier to find the answer Using the formula? Using the formula? Using common sense? Using common sense?

3. Conditional Probability Suppose that for a randomly selected day, Suppose that for a randomly selected day, P(wind) = 0.3. P(wind) = 0.3. P(rain) = 0.2. P(rain) = 0.2. P(wind | rain) = 0.4. P(wind | rain) = 0.4. Find P(rain and wind). Find P(rain and wind). Find P(rain | wind). Find P(rain | wind).

4. Independent Events See Example 7.5, p. 402 – Gender versus Education. See Example 7.5, p. 402 – Gender versus Education. Let Let A = Person is female. A = Person is female. B = Person has college degree. B = Person has college degree. Find Find P(A and B) by using the table. P(A and B) by using the table. P(A and B) by using the formula. P(A and B) by using the formula.

5. Independent Events See the example on pages 200 – 201 in Chapter 4. See the example on pages 200 – 201 in Chapter 4. The chart on p. 200 shows the marginal distribution of nutritional status The chart on p. 200 shows the marginal distribution of nutritional status The chart on p. 201 shows the conditional distributions of nutritional status, given academic performance. The chart on p. 201 shows the conditional distributions of nutritional status, given academic performance. Based on the chart, are academic performance and nutritional status independent? Based on the chart, are academic performance and nutritional status independent?

6. Let’s Do It! Let’s do it! 7.15, p. 404 – A Family Plan Revisited. Let’s do it! 7.15, p. 404 – A Family Plan Revisited. Let’s do it! 7.16, p. 404 – Getting Good Business? Let’s do it! 7.16, p. 404 – Getting Good Business?

7. Example Suppose that for a randomly selected day, Suppose that for a randomly selected day, P(wind) = 0.3. P(wind) = 0.3. P(rain) = 0.2. P(rain) = 0.2. P(wind and rain) = 0.06. P(wind and rain) = 0.06. Find P(wind | rain) and P(rain | wind). Find P(wind | rain) and P(rain | wind). Are the events “wind” and “rain” independent? Are the events “wind” and “rain” independent?

8. Repeated Independent Trials Suppose that 48% of the population think the Iraq War was a bad idea and 52% think it was a good idea. Suppose that 48% of the population think the Iraq War was a bad idea and 52% think it was a good idea. If we select 3 people at random, what is the probability that all three of them think the war was a good idea? If we select 3 people at random, what is the probability that all three of them think the war was a good idea?

9. Repeated Independent Trials: All Each has probability 0.52 of thinking the war was a good idea. Each has probability 0.52 of thinking the war was a good idea. Since the trials are independent, then the probability that all three believe so is Since the trials are independent, then the probability that all three believe so is (0.52)(0.52)(0.52) = (0.52) 3 = 0.1406.

10. Repeated Independent Trials: None What is the probability that none of them believe that it was a good idea? What is the probability that none of them believe that it was a good idea? That’s the same as the probability that all of them believe it was a bad idea. That’s the same as the probability that all of them believe it was a bad idea. The probability that one individual believes it was a bad idea is 0.48. The probability that one individual believes it was a bad idea is 0.48. (0.48) 3 = 0.1106.

11. Repeated Independent Trials: At Least One What is the probability that at least one believes that it was a good idea? What is the probability that at least one believes that it was a good idea? The complement of “at least one” is “none.” The complement of “at least one” is “none.” Therefore, the answer is Therefore, the answer is 1 – (0.48) 3 = 1 – 0.1106 = 0.8894.

12. Repeated Independent Trials: Exactly Two What is the probability that exactly two of them believe that it was a good idea? What is the probability that exactly two of them believe that it was a good idea? The question does not specify which two, so we must include all possibilities. The question does not specify which two, so we must include all possibilities. 1 st and 2 nd (only). 1 st and 2 nd (only). 1 st and 3 rd (only). 1 st and 3 rd (only). 2 nd and 3 rd (only). 2 nd and 3 rd (only).

13. Repeated Independent Trials: Exactly Two P(exactly two) = P((1 st and 2 nd only) or (1 st and 3 rd only) or (2 nd and 3 rd only)). P(exactly two) = P((1 st and 2 nd only) or (1 st and 3 rd only) or (2 nd and 3 rd only)). At this point, it is helpful to draw and label a tree diagram. At this point, it is helpful to draw and label a tree diagram. The possibilities are GGB, GBG, and BGG. The possibilities are GGB, GBG, and BGG.

14. Repeated Independent Trials: Exactly Two P(exactly two) = P(GGB) + P(GBG) P(exactly two) = P(GGB) + P(GBG) + P(BGG) = (.52)(.52)(.48) + (.52)(.48)(.52) + (.48)(.52)(.52) = 0.3894.

15. Assignment Page 406: Exercises 27 - 33, 37. Page 406: Exercises 27 - 33, 37. Page 451: Exercises 75*, 80, 82, 86, 87, 89, 90. Page 451: Exercises 75*, 80, 82, 86, 87, 89, 90. * Assume that no one knows the number that anyone else is choosing.