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Poisson Distribution: A small life insurance company has determined that on the average it receives 3 death claims per day. Find the probability that the company receives at least four death claims on a randomly selected day.
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for mean m \[P(n=i) = \frac{e^{-m} m^i} {i!}\] we have m = 3, we want P(i > 4) a bit of a pain \[e^{-3} \approx 0.05\] P(n=0) = 0.05 P(n=1) = 0.05 x 3 / 1 P(n=2) = 0.05 x 9 / 2 P(n=3) = 0.05 x 27 / 6
I forgot to include n=4. Calculate the five terms, add them together, and subtract them from 1.
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