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Unformatted text preview: ORIE 3500/5500, Fall 10 HW 2 Solutions Assignment 2 Solutions Problem 1 Let A be the event that a type i coupon is not chosen in the set of k coupons and B be the event that a type j coupon is not chosen in the set of k coupons. Then, P ( A ) = (1- p i ) k , P ( B ) = (1- p j ) k , P ( A B ) = (1- p i- p j ) k . So, the event that a type i coupon is chosen but not a type j coupon is chosen is A c B and P ( A c B ) = P ( B )- P ( A B ) = (1- p j ) k- (1- p i- p j ) k . Problem 2 (a) By the assumption A c B = . Hence P ( B ) = P ( A B ) + P ( A c B ) = P ( A B ) . From the inclusion-exclusion formula it follows that P ( A B ) = P ( A ) + P ( B )- P ( A B ) = P ( A ) = 0 . 7 . (b) If A c and B are independent, so are A and B , since P ( A ) P ( B ) = (1- P ( A c )) P ( B ) = P ( B )- P ( A c B ) = P ( A B ). This result is also obtained by the following fact: if events A 1 ,A 2 ,...,A n are independent, then so are any derived events constructed from disjoint groupings of the events A 1 ,A 2 ,...,A n (See the lecture note on Aug 30). Therefore P ( A B ) = P ( A ) + P ( B )- P ( A B ) = 1 .....
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This note was uploaded on 10/29/2011 for the course MATH 3310 at Cornell University (Engineering School).