Question

The number of accidents in a production facility has a Poisson distribution with a mean of 2.8 per month. For a given month, what is the probability that there will be more than three (3) accidents?

Answer 1

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

$=1-e^{-2.8}(\frac{2.8^0}{0!}+\frac{2.8^1}{1!}+\frac{2.8^2}{2!}+\frac{2.8^3}{3!})=0.3081.$