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Unformatted text preview: IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 6 Solution Chapter 4 Question 51 (a) e . 2 (b) 1 e . 2 . 2 e . 2 = 1 1 . 2 e . 2 Since each letter has small probability of being a typo, the number of errors should approximately have a Poisson distribution. Question 53 (a) P { both born on April 30 } = 1 365 2 , then P { at least one couple both born on April 30 } = 1 P { no couple both born on Apri } = 1 exp 80000 365 2 = . 451 (b) P { both born on the same day } = 1 365 , then P { at least one couple both born on the same day } = 1 P { no couple both born on the same day } = 1 exp 80000 365 1 Each couple has a small probability of having the same birthday, so the number of couples having the same birthday should approximately have a Poisson distribution. Question 58 (a) Binomial value B (2 , 8 ,. 1) = . 1488 Poisson approximation value Poisson(2 , = . 8) = . 1438 ....
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This note was uploaded on 04/02/2008 for the course IEOR 172 taught by Professor Righter during the Fall '07 term at University of California, Berkeley.
 Fall '07
 righter

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