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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 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- . 99 . Therefore, e- . 01 , e 100 , 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 300-page 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 = 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 10-second period? What assumptions are you making? Solution. Let X be the number of drops hitting a particular square inch in a given 10-second period. If we assume that the drops are hitting according to a Poisson distribution, then the average number of drops in a 10-second period is one sixth of the average number of drops per minute, or = 5. Hence P ( X = 0) = e-...
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- Fall '03