Let X be a random variable which indicates number of cars per minute.
Given E(X) (i.e) expected number of cars per minute (i.e) average number of cars per minute (i.e) mean number of cars per minute = 3
probability of number of cars being 0 or 1 or 2 per minute
= P(X = 0) + P(X = 1) + P(X = 2)
= e^{-3} (3)^{0} / 0! + e^{-3} (3)^{1} / 1! + e^{-3} (3)^{2} / 2!
= 17 / 2e^{3 }