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

We have a Poisson distribution,

$\lambda=2.7;t=1;\lambda t=2.7\cdot1=2.7;\\ P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!}=\cfrac{2.7^k\cdot e^{-2.7}}{k!}.$

$P_{1}(X>3)=1-P(X\le3)=\\ =1-(P(X=0)+P(X=1)+\\+P(X=2)+P(X=3));\\ P_1(X=0)=\cfrac{2.7^0\cdot e^{-2.7}}{0!}=0.0672;\\ P_1(X=1)=\cfrac{2.7^1\cdot e^{-2.7}}{1!}=0.1815;\\ P_1(X=2)=\cfrac{2.7^2\cdot e^{-2.7}}{2!}=0.2450;\\ P_1(X=3)=\cfrac{2.7^3\cdot e^{-2.7}}{3!}=0.2205;\\ P_1(X>3)=1-(0.0672+0.1815+\\ +0.2450+0.2205)=0.2859.$