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Unformatted text preview: Stat 116 Homework 2 Solutions April 9, 2008 1. Let PSA + = “test indicates elevated PSA” PSA − = “test indicates not elevated PSA” C = “has cancer” NC = “does not have cancer” We know that P ( PSA +  NC ) = 0 . 135, P ( PSA +  C ) = 0 . 268 and P ( C ) = 0 . 7. (a) P ( C  PSA +) = P ( PSA +  C ) P ( C ) P ( PSA +) = P ( PSA +  C ) P ( C ) P ( PSA +  C ) P ( C ) + P ( PSA +  NC ) P ( NC ) = . 268 × . 7 . 268 × . 7 + 0 . 135 × . 3 = 0 . 822 (b) P ( C  PSA − ) = P ( PSA −  C ) P ( C ) P ( PSA − ) = P ( PSA −  C ) P ( C ) P ( PSA −  C ) P ( C ) + P ( PSA −  NC ) P ( NC ) = . 732 × . 7 . 732 × . 7 + 0 . 865 × . 3 = 0 . 664 2. For any i negationslash = j , P ( A i,j ) = # ways 2 people have same BD # ways 2 people have BDs = 365 365 2 = 1 365 . To show that A i,j are pairwise independent, we only need to consider A i,j ∩ A j,k , where i negationslash = j negationslash = k , and A i,j ∩ A r,s , where i negationslash = j negationslash = r negationslash = s . Now P ( A i,j ∩ A j,k ) = # of ways 3 people have same BD # ways 3 people have BDs = 365 365 3 = 1 365 2 = P ( A i,j ) P ( A j,k ) 1 and P ( A i,j ∩ A r,s ) = # of ways i and j have same BD...
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This note was uploaded on 01/13/2010 for the course STATS 116 taught by Professor Staff during the Spring '07 term at Stanford.
 Spring '07
 Staff
 Probability

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