Answer
$P(4)=0.195$
On about 195 of every 1000 days, there will be exactly 4 cars arriving between 4:00 p.m. and 4:10 p.m.
Work Step by Step
$P(x)=\frac{(λt)^x}{x!}e^{-λt}$
λ = 0.41 -> (0.41 car every minute)
t = 10 -> ($4:10~P.M.-4:00~P.M.=10~min$)
$P(4)=\frac{(0.41\times10)^4}{4!}e^{-0.41\times10}=\frac{282.5761}{24}e^{-4.1}=0.195$
