In a town 10 accidents took place in a period of 50 days. Assume that the number of accidents per day follow Poisson distribution. Find the probability that there will be three or more accidents per day.
For random variable from Poisson distribution with parameter "\\lambda" :
"\\lambda = EX = \\frac{10}{50} = 0.2" , where "EX" is sample mean
Distribution function:
"Pr(X=k) = e^{-\\lambda}\\frac{\\lambda^k}{k!} = e^{-0.2}\\frac{0.2^k}{k!}"
Hence,
"Pr(X \\geq 3) = 1 - Pr(X < 3) = 1 - \\sum_{k=0}^{2}Pr(X=k) = 1 - e^{-0.2}(1 + \\frac{0.2}{1} + \\frac{0.2^2}{2}) = 0.0011"
So, the probability is approximately 0.11%
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