The number of accidents in a production facility has a Poisson distribution with a mean of 2.7 per month.For a given month what is the probability that there will be more than three accidents?
Given mean "\\lambda=2.7,"
Let "X" denote the number of accidents-
Probability that there will be more than three accidents-
"P(X>3)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)]"
"=1-[\\dfrac{e^{-\\lambda}.\\lambda^0}{0!}+\\dfrac{e^{-\\lambda}.\\lambda^1}{1!}+\\dfrac{e^{-\\lambda}.\\lambda^2}{2!}+\\dfrac{e^{-\\lambda}.\\lambda^3}{3!}]"
"=1-[\\dfrac{e^{-2.7}.(2.7)^0}{1}+\\dfrac{e^{-2.7}.(2.7)^1}{1}+\\dfrac{e^{-2.7}.(2.7)^2}{2}+\\dfrac{e^{-2.7}.(2.7)^2}{6}]\\\\[9pt]=1-(0.0672+0.08145+0.24496+0.22046)\\\\[9pt]=1-0.61767\\\\[9pt]=0.38233"
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