Question

Telephone calls are being placed through a certain exchange at random times on the

average of four per minute. Assuming a Poisson process, determine the probability

that in a 15-second intervals, there are 3 or more calls.

Answer 1

$P(X=x) = \frac{e^{-λ}λ^x}{x!} \\ λ=\frac{4}{4}=1 \\ P(X≥3) = 1- P(X\leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)] \\ = 1- [\frac{e^{-1}(1)^0}{0!}+\frac{e^{-1}(1)^1}{1!}+ \frac{e^{-1}(1)^2}{2!}] \\ = 1-e^{-1}(1+1+\frac{1}{2}) \\ =1-2.5 \times e^{-1} \\ =1-0.9175 \\ =0.0825$