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I have a problem where I am given that accidents occur at a Poisson rate of 3 per day. I am asked to find the probability that in January (31 days long), exactly 3 days will be accident-free. The correct way to do this is to find a Poisson rate $P(N(1)=0),$ so the probability of 0 accidents per 1 day using $P(N(1)=0) = \frac{(3\cdot1)^0 e^{-3}}{0!} = 0.0498$ (since $\lambda = 3$), and then do $\binom{31}{3} (0.0498)^3 (1-0.0498)^{28} = 0.132759$ Intuitively, this answer ring the bell to me, but my initial approach was the following: find $P(N(3)=0),$ or a Poisson process estimation of probability that in 3 days, 0 accidents happen, and count $\binom{31}{3} P(N(3) = 0) = \binom{31}{3} \frac{(3\cdot3)^{0} e^{-9}}{0!} = 0.555.$ This is clearly way off, and I would be happy if someone could explain me what is it that I found in the second case and why it makes no sense.

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    @DilipSarwate: Yes, I can see how my second remark would compound the problem, when read that way. I fear I was a bit too flippant on both counts. :)2011-10-13

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