Question

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?

Answer 1

Given mean $\lambda=2.7,$

Let $X$ denote the number of accidents-

 Probability that there will be more than three accidents-

$P(X>3)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$=1-[\dfrac{e^{-\lambda}.\lambda^0}{0!}+\dfrac{e^{-\lambda}.\lambda^1}{1!}+\dfrac{e^{-\lambda}.\lambda^2}{2!}+\dfrac{e^{-\lambda}.\lambda^3}{3!}]$

$=1-[\dfrac{e^{-2.7}.(2.7)^0}{1}+\dfrac{e^{-2.7}.(2.7)^1}{1}+\dfrac{e^{-2.7}.(2.7)^2}{2}+\dfrac{e^{-2.7}.(2.7)^2}{6}]\\[9pt]=1-(0.0672+0.08145+0.24496+0.22046)\\[9pt]=1-0.61767\\[9pt]=0.38233$