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Conditional Probability. Lecture 33 Section 7.4.2 Tue, Mar 23, 2004. Conditional Probability. Suppose that for a randomly selected day, P(wind) = 0.3. P(rain) = 0.2. P(wind and rain) = 0.1. Find P(wind | rain) and P(rain | wind). Is it easier to find the answer Using the formula?
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Conditional Probability Lecture 33 Section 7.4.2 Tue, Mar 23, 2004
Conditional Probability • Suppose that for a randomly selected day, • P(wind) = 0.3. • P(rain) = 0.2. • P(wind and rain) = 0.1. • Find P(wind | rain) and P(rain | wind). • Is it easier to find the answer • Using the formula? • Using common sense?
Conditional Probability • Suppose that for a randomly selected day, • P(wind) = 0.3. • P(rain) = 0.2. • P(wind | rain) = 0.4. • Find P(rain and wind). • Find P(rain | wind).
Independent Events • See Example 7.5, p. 402 – Gender versus Education. • Let • A = Person is female. • B = Person has college degree. • Find • P(A and B) by using the table. • P(A and B) by using the formula.
Independent Events • 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. 201 shows the conditional distributions of nutritional status, given academic performance. • Based on the chart, are academic performance and nutritional status independent?
Let’s Do It! • Let’s do it! 7.15, p. 404 – A Family Plan Revisited. • Let’s do it! 7.16, p. 404 – Getting Good Business?
Example • Suppose that for a randomly selected day, • P(wind) = 0.3. • P(rain) = 0.2. • P(wind and rain) = 0.06. • Find P(wind | rain) and P(rain | wind). • Are the events “wind” and “rain” independent?
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. • If we select 3 people at random, what is the probability that all three of them think the war was a good idea?
Repeated Independent Trials: All • 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 (0.52)(0.52)(0.52) = (0.52)3 = 0.1406.
Repeated Independent Trials: None • 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. • The probability that one individual believes it was a bad idea is 0.48. (0.48)3 = 0.1106.
Repeated Independent Trials: At Least One • What is the probability that at least one believes that it was a good idea? • The complement of “at least one” is “none.” • Therefore, the answer is 1 – (0.48)3 = 1 – 0.1106 = 0.8894.
Repeated Independent Trials: Exactly Two • 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. • 1st and 2nd (only). • 1st and 3rd (only). • 2nd and 3rd (only).
Repeated Independent Trials: Exactly Two • P(exactly two) = P((1st and 2nd only) or (1st and 3rd only) or (2nd and 3rd only)). • At this point, it is helpful to draw and label a tree diagram. • The possibilities are GGB, GBG, and BGG.
Repeated Independent Trials: Exactly Two • P(exactly two) = P(GGB) + P(GBG) + P(BGG) = (.52)(.52)(.48) + (.52)(.48)(.52) + (.48)(.52)(.52) = 0.3894.
Assignment • Page 406: Exercises 27 - 33, 37. • Page 451: Exercises 75*, 80, 82, 86, 87, 89, 90. * Assume that no one knows the number that anyone else is choosing.
