Independent solution

How to solve this Discrete Random Variables question

Setup

Setup

Let N be the count in the stated one-year interval. Its Poisson mean is two.

N∼Poisson⁡(2),Pr⁡(N=k)=e−22kk!N\sim\operatorname{Poisson}(2),\qquad \Pr(N=k)=e^{-2}\frac{2^k}{k!}

Model

Model

Not exceeding four means that the possible counts are zero, one, two, three, and four.

Pr⁡(N≤4)=∑k=04e−22kk!\Pr(N\le4)=\sum_{k=0}^{4}e^{-2}\frac{2^k}{k!}

Compute

Compute

Factor out the common exponential term and add the five finite-series coefficients.

Pr⁡(N≤4)=e−2(1+2+2+43+23)\Pr(N\le4)=e^{-2}\left(1+2+2+\frac{4}{3}+\frac{2}{3}\right)
Pr⁡(N≤4)=7e−2=0.9473469827\Pr(N\le4)=7e^{-2}=0.9473469827

Answer

Answer

The cumulative probability through count four is 7 exp(-2).

7e−2(E)\boxed{7e^{-2}\quad\text{(E)}}