Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let X and Y be the two annual event counts. The required outcome occurs when each count stays within its own limit.

X∼Poisson⁡(1),Y∼Poisson⁡(0.3)X\sim\operatorname{Poisson}(1),\qquad Y\sim\operatorname{Poisson}(0.3)
{both limits met}={X≤2}∩{Y≤1}\{\text{both limits met}\}=\{X\le2\}\cap\{Y\le1\}

Model

Model

Independence allows the joint event probability to factor into two Poisson cumulative probabilities.

Pr⁡(X≤2,Y≤1)=Pr⁡(X≤2)Pr⁡(Y≤1)\Pr(X\le2,Y\le1)=\Pr(X\le2)\Pr(Y\le1)

Compute

Compute

Sum each Poisson mass from zero through the applicable upper bound.

Pr⁡(X≤2)=e−1(1+1+12)=2.5e−1\Pr(X\le2)=e^{-1}\left(1+1+\frac{1}{2}\right)=2.5e^{-1}
Pr⁡(Y≤1)=e−0.3(1+0.3)=1.3e−0.3\Pr(Y\le1)=e^{-0.3}(1+0.3)=1.3e^{-0.3}
Pr⁡(X≤2,Y≤1)=(2.5e−1)(1.3e−0.3)=0.8857283274…\Pr(X\le2,Y\le1)=(2.5e^{-1})(1.3e^{-0.3})=0.8857283274\ldots

Answer

Answer

The product matches the expression in choice B.

(2.5e−1)(1.3e−0.3)(B)\boxed{(2.5e^{-1})(1.3e^{-0.3})\quad\text{(B)}}