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Unformatted text preview: Math 4653: Elementary Probability: Spring 2007 Homework #5. Problems and Solutions 1. Sec. 3.5: #2: How many raisins must cookies contain on average for the chance of a cookie containing at least one raisin to be at least 99%? Solution. Let X be the number of raisins in a particular cookie, and let μ be the average number of raisins per cookie. If we assume that the raisins are distributed to the cookies through a Poisson process, then P ( X ≥ 1) = 1 P ( X = 0) = 1 e μ ≥ . 99 . Therefore, e μ ≤ . 01 , e μ ≥ 100 , μ ≥ ln 100 ≈ 4 . 6 , so the average should be at least 4.6 raisins per cookie. 2. Sec. 3.5: #4: Books from a certain publisher contain an average of 1 misprint per page. What is the probability that on at least one page in a 300page book from this publisher there will be at least 5 misprints? Solution. Let X be the number of misprints on a particular page. We are given that the average number of misprints per page is μ = 1. If we assume that the misprints are distributed to the pages through a Poisson process, then P ( X ≥ 5) = 1 4 X k =0 P ( X = k ) = 1 e 1 4 X k =0 1 k ! = 1 e 1 65 24 ≈ . 003660 . In other words, the probability of a particular page containing at least 5 misprints is p . Therefore, if E is the event that at least one page in the book contains at least 5 misprints, then E c is the event that no page in the book contains at least 5 misprints, so P ( E c ) = (1 p ) 300 = e 1 65 24 300 . Therefore, P ( E ) = 1 P ( E c ) = 1 e 1 65 24 300 ≈ . 66712 . 3. Sec. 3.5: #6: Suppose rain is falling at an average rate of 30 drops per square inch per minute. What is the chance that a particular square inch is not hit by any drops during a given 10second period? What assumptions are you making? Solution. Let X be the number of drops hitting a particular square inch in a given 10second period. If we assume that the drops are hitting according to a Poisson distribution, then the average number of drops in a 10second period is one sixth of the average number of drops per minute, or μ = 5. Hence P ( X = 0) = e...
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This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at Berkeley.
 Fall '03
 aldous
 Probability

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