This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
This note was uploaded on 04/07/2011 for the course MATH 302 taught by Professor Israel during the Spring '08 term at The University of British Columbia.
 Spring '08
 ISRAEL
 Probability

Click to edit the document details