Question

. Assume that arrivals occur according to a Poisson process with an average

of seven per hour. What is the probability that exactly two customers arrive in the two-hour period

of time between

(a) 2:00 P.M. and 4:00 P.M. (one continuous two-hour period)?

(b) 1:00 P.M. and 2:00 P.M. or between 3:00 P.M. and 4:00 P.M. (two separate one-hour periods

that total two hours)?

Answer 1

We have a Poisson distribution,

$\lambda=7;\\ P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!}.$

$(a) \ \ P_{2}(X=2)=\cfrac{(7\cdot2)^2\cdot e^{-7\cdot2}}{2!}=\\ =0.00008.$

$(b) \ \ P_{1}(X=2)\cdot P_{1}(X=2)=\\ =\begin{pmatrix} \cfrac{(7\cdot1)^2\cdot e^{-7\cdot1}}{2!} \end{pmatrix}^2=\\ =0.02234^2=0.00050.$