Question

If 1.6 accidents can be expected an intersection on any given day, what is the

probability that there will be 3 accidents on any given day?

Answer 1

We have a Poisson distribution,

$λ=1.6;\\ P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!}=\cfrac{(1.6t)^k\cdot e^{-1.6t}}{k!},\\ P_1(X=3)=\cfrac{(1.6\cdot1)^3\cdot e^{-1.6\cdot1}}{3!}=0.1378.$