IE230 fall2015 hw1solutions PDF

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IE230 fall2015 hw1solutions...

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Solutions to Homework 1 (IE 230, Fall 2011) Notes: 1. The question numbers correspond to the 4th edition of the textbook. You can refer to the original homework assignment to find out the corresponding numbers for the 5th . 2. The solutions may or may not contain enough intermediate steps to arrive at the final answers. In your homework solutions (and quizzes and exams for that matter), you should try to provide as many necessary details as you can to earn partial credits.

!2#8.!Let!an!ordered!pair!of!numbers,!such!as!43!denote!the!response!on!the!first!! !!!!!!!!!and!second!question.!Then,!S!consists!of!the!25!ordered!pairs!{11,!12,!…,!55}.! ! 2-14. automatic transmission

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2-28.

A c  B = 55, Bc =23, A  B = 85

2-50.

All outcomes are equally likely a) P(A) = 2/5 b) P(B) = 3/5 c) P(A') = 3/5 d) P(AB) = 1 e) P(AB) = P()= 0

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! !2#65!! c) Now, B = A  ( A c  B) and the events A and A c  B are mutually exclusive. Therefore, P(B) = P(A) + P( A c  B ). Because P( A c  B ) t 0 , P(B) t P(A). ! ! !

2-66.

a) P(A') = 1- P(A) = 0.7 b) P ( A  B ) = P(A) + P(B) - P( A B ) = 0.3+0.2 - 0.1 = 0.4 c) P( A c  B ) + P( A  B ) = P(B). Therefore, P( A c B ) = 0.2 - 0.1 = 0.1 d) P(A) = P( A  B ) + P( A  B c ). Therefore, P( A  B c ) = 0.3 - 0.1 = 0.2 e) P(( A  B )') = 1 - P( A  B ) = 1 - 0.4 = 0.6 f) P( A c  B ) = P(A') + P(B) - P( A c  B ) = 0.7 + 0.2 - 0.1 = 0.8 !...