Question

1. 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 (3) accidents?

Answer 1

$P(X>3)=1-P(X\le3)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))=$

$=1-e^{-2.7}(\frac{2.7^0}{0!}+\frac{2.7^1}{1!}+\frac{2.7^2}{2!}+\frac{2.7^3}{3!})=0.2859.$