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# s1 - Stat 5101(Geyer 2002 Midterm 1 Problem 1 This is like...

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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 non-defective 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 ) = 0 . 96 × 0 . 05 0 . 96 × 0 . 05 + 0 . 02 × 0 . 95 = 0 . 7164 Problem 3 This is like Problem 3.2.4 in DeGroot and Schervish. 1

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We need to establish 1 = Z 1 - 1 f ( x ) dx = Z 1 - 1 c (1 - x 2 ) dx = c x - x 3 3 ¶fl fl fl fl 1 - 1 = 4 c 3 Hence c = 3 4 Problem 4 This is like Additional Problem 2 on the homework.
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