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Unformatted text preview: Math 302.102 Fall 2010 Solutions to Assignment #5 1. (a) Suppose that A is the event A = { John smoked marijuana } , and suppose further that B is the event B = { John said yes } . The law of total probability implies that P { B } = P { B  A } P { A } + P { B  A c } P { A c } . We do not know whether or not John said yes. We only know that 40 / 100 = 0 . 40 of respondents said yes. Therefore, P { B } = 0 . 4. We also know that if John has, in fact, smoked marijuana, then he would have answered truthfully (i.e., said yes) provided that he rolled a 1, 2, 3, or 4. That is, P { B  A } = 4 / 6. Similarly, if we know that John has not smoked marijuana, then he would have lied (i.e., said yes) provided that he rolled a 5 or 6. That is, P { B  A c } = 2 / 6. In other words, we find . 4 = 4 6 P { A } + 2 6 P { A c } . Of course, P { A c } = 1 P { A } implying that . 4 = 4 6 P { A } + 2 6 (1 P { A } ) and so P { John smoked marijuana } = P { A } = . 4 2 / 6 4 / 6 2 / 6 = 0 . 2 . (b) By Bayes’ rule, the required probability is P { John smoked marijuana  John said yes } = P { A  B } = P { B  A } P { A } P { B } = 4 / 6 · . 2 . 4 = 1 3 . 2. Let W j , R j , B j denote the event that a white ball, red ball, black ball, respectively, was selected on draw j . (a) Using the law of total probability we find P { B 2 } = P { B 2  W 1 } P { W 1 } + P { B 2  R 1 } P { R 1 } + P { B 2  B 1 } P { B 1 } = b w + r + b + d · w w + r + b + b w + r + b + d · r w + r + b + b + d w + r + b + d · b w + r + b = 1 w + r + b + d · 1 w + r + b [ bw + br + ( b + d ) b ] = 1 w + r + b + d · 1 w + r + b · b [ w + r + b + d ] = b w + r + b . Notice that the probability that the second ball is black is the same as the probability that the first ball is black; that is, P { B 1 } = P { B 2 } = b/ ( w + r + b ). However, the events B 1 and B 2 are not independent. This is because P { B 1 ∩ B 2 } = P { B 2  B 1 } P { B 1 } = b + d w + r...
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 Spring '08
 ISRAEL
 Conditional Probability, Probability, Probability theory, dx

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