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Unformatted text preview: January. N ( t ) is a Poisson process with rate 3. Therefore, P ( N (1) = 0) = e3 ≈ . 0498 . By the independence of increments of Poisson processes, the probability of no accidents on any given day is equal to 0.0498. Therefore, if Y gives the number of days in January for which there are no accidents, Y is binomial with parameters n = 31 and p = 0 . 0498. Therefore, P ( Y = 3) = p 31 3 P . 0498 3 (1. 0498) 2 8 = 31 * 30 * 29 6 . 0498 3 . 9502 2 8 = 0 . 1328 . 1 Another way to do this is to consider Y as approximately Poisson. This can be (barely) justifed because n is moderately large, p is small and np = 31 * . 0498 = 1 . 5438. Thus, Y is approximately Poisson with λ = 1 . 5438 and so P ( Y = 3) ≈ e1 . 5438 (1 . 5438) 3 / 6 ≈ . 1310 . Note that these answers are mildly diFerent. 2...
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 Spring '08
 Anderson
 Poisson Distribution, Probability, Probability theory, Immigration to the United States, Poisson process, E2

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