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 (3) accidents?

Answer 1

X~Pois(2.7)

$P(X>3)=1-P(X≤3)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)=1-e^{-2.7}({\frac {2.7^0} {0!}}+{\frac {2.7^1} {1!}}+{\frac {2.7^2} {2!}}+{\frac {2.7^3} {3!}})\approx0.286$