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

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Unformatted text preview: 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 ) = . 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 ) . 07 = q + q- . 01 . 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- . 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 ) . 07 = Pr ( A 2 | A 1 ) Pr ( A 1 ) + Pr ( A 1 | A 2 ) Pr ( A 2 ) + 0 . 01 . 06 = (1- r ) q + (1- r ) q . 06 = 2 1- . 01 q q . 06 = 2 q- . 02 . 08 = 2 q . 04 = q 1 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 (...
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hw3 solutions - Solutions to Homework Assignment#3...

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