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Unformatted text preview: Stat 5101 (Geyer) 2002 Midterm 1 Problem 1 This is like Problems 1.8.4 and 1.8.10 in DeGroot and Schervish. There are ( 20 5 ) ways to choose 5 light bulbs out of 20 light bulbs. So that is the size of the sample space. The phrase selected at random in the problem statement means all are equally likely. There are ( 17 5 ) ways to choose 5 light bulbs out of the 17 nondefective light bulbs. Hence the probability is ( 17 5 ) ( 20 5 ) = 6188 15504 = 0 . 3991 Problem 2 This is a Bayes rule problem like Problem 2.3.8 in DeGroot and Schervish. Let A be the event tests positive and B be the event uses drugs. Then we are given Pr( A  B ) = 0 . 96 Pr( A c  B ) = 0 . 04 Pr( A  B c ) = 0 . 02 Pr( A c  B c ) = 0 . 98 Pr( B ) = 0 . 05 from which we can derive (by the complement rule) Pr( B c ) = 0 . 95 What we want is Pr( B  A ) = Pr( A  B )Pr( B ) Pr( A  B )Pr( B ) + Pr( A  B c )Pr( B c ) = . 96 . 05 . 96 . 05 + 0 . 02 . 95 = 0 . 7164 Problem 3 This is like Problem 3.2.4 in DeGroot and Schervish....
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This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Fall '02 term at Minnesota.
 Fall '02
 Staff

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