Question

A hospital switch board receives on an average 0.9 calls/minute. Find the probability that (i) no calls in a minute, (ii) 2 or more calls/minute

Answer 1

Let's use Poisson's distribution $P(x=k)=\lambda^ke^{-\lambda}/k!,$

where $\lambda$ is the mean

$1)\space P(x=0)=0.9^0e^{-0.9}/0!=\\ =1/e^{0.9}\approx0.4066=40.66\%\\ 2)\space P(x\geq2)=1-P(x<2)=\\ =1-P(x=0)-P(x=1)=\\ =1-0.4066-0.9^1e^{-0.9}/1!\approx0.2275=22.75\%$