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Unformatted text preview: Homework 2  KEY Jeff Brenion June 12, 2004 Note: Many problems can be solved in more than one way; we present only a single solution here. 1 Problem 14 The problem statement gives us the probabilities of events A , B , and C , as well as their various intersections (for example, P ( A ∩ B ) is listed as 0 . 5.) By Theorem 14, we can solve this problem through inclusionexclusion: P ( A ∪ B ∪ C ) = . 02 + . 01 + . 015 . 005 . 006 . 004 + . 002 = .032 2 Problem 115 Note that there are 300 p defecctive units and 300(1 p ) functional units. We can break this problem into two cases: the case where no defective items are found in the sample, and the case where one defective item is found in the sample. Since all of the possible samples are equally likely, this reduces to a counting problem; there are ( 300(1 p ) 10 ) ways for the first case to happen, and ( 300(1 p ) 9 )( 300 p 1 ) ways for the second, as well as ( 300 10 ) different ways to select a sample without regard to the number of defective units. Thus, theselect a sample without regard to the number of defective units....
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This note was uploaded on 09/25/2011 for the course ECON 101 taught by Professor Wang during the Spring '11 term at SUNY Buffalo.
 Spring '11
 wang

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