hw3 solutions

# Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Solutions to Homework Assignment #3 Statistics 220, Fall 2005 Due September 21, 2005 2.52 Let A 1 = “pump #1 fails”, and A 2 = “pump #2 fails”. Because the pumps are identical, we can assume that Pr ( A 1 ) = Pr ( A 2 ) = q . The problem requires us to find q . The easy way to solve this is to use the fact that we’re given Pr ( A 1 A 2 ) = 0 . 01, the probability of both pumps failing, and Pr ( A 1 A 2 ) = 0 . 07, the prob- ability of at least one failure. So Pr ( A 1 A 2 ) = Pr ( A 1 ) + Pr ( A 2 ) + Pr ( A 1 A 2 ) 0 . 07 = q + q - 0 . 01 0 . 08 = 2 q q = 0 . 04 Notice that this doesn’t use the fact, which we’re also given, that r = Pr ( A 2 | A 1 ) > q . A slightly more complicated approach does. rq = Pr ( A 2 | A 1 ) Pr ( A 1 ) = Pr ( A 1 A 2 ) = 0 . 01, so r = 0 . 01 /q . Similarly, Pr ( A 1 | A 2 ) = 0 . 01 /q = r . And Pr ( A 2 | A 1 ) = 1 - r = 1 - 0 . 01 q = Pr ( A 1 | A 2 ). Pr ( A 1 A 2 ) = Pr ( A 1 A 2 ) + Pr ( A 2 A 1 ) + Pr ( A 1 A 2 ) 0 . 07 = Pr ( A 2 | A 1 ) Pr ( A 1 ) + Pr ( A 1 | A 2 ) Pr ( A 2 ) + 0 . 01 0 . 06 = (1 - r ) q + (1 - r ) q 0 . 06 = 2 1 - 0 . 01 q q 0 . 06 = 2 q - 0 . 02 0 . 08 = 2 q 0 . 04 = q 1

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2.58 We did this one in class: Pr ( A B | C ) = Pr (( A B ) C ) Pr ( C ) = Pr (( A C ) ( B C )) Pr ( C ) = Pr ( A C ) + Pr ( B C ) - Pr (( A C ) ( B C )) Pr ( C ) = Pr ( A | C ) + Pr ( B | C ) - Pr ( A B C ) Pr ( C ) = Pr ( A | C ) + Pr ( B | C ) - Pr ( A B | C ) The restriction Pr ( C ) > 0 is just so that the denimonator makes sense. 2.64 Looking at example 2.30 on p. 82, we have A 1 = “has the disease”, A 2 = “does not have the disease”, and B = “positive test result”. Pr ( B | A 1 ) = 0 . 99, and Pr ( B | A 2 ) = 0 . 02. We’re told in the problem that Pr ( A 1 ) = 1 25 = 0 . 04, so Pr ( A 2 ) = 1 - Pr ( A 1 ) = 0 . 96. The probability of a positive result comes from the rule of total probability: Pr ( B ) = Pr ( B | A 1 ) Pr ( A 1 ) + Pr ( B | A 2 ) Pr ( A 2 ) = 0 . 059 The probability of having the disease given a positive result comes from Bayes’s rule: Pr ( A 1 | B ) = Pr ( B | A 1 ) Pr ( A 1 ) Pr ( B ) = 0 . 67
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