Independent solution

How to solve this Poisson Aggregation question

Setup

Setup

The three independent weekly Poisson counts combine into one Poisson count with mean equal to the sum of the weekly means.

T∼Poisson⁡(3⋅2)=Poisson⁡(6)T\sim\operatorname{Poisson}(3\cdot2)=\operatorname{Poisson}(6)

Model

Model

The aggregate mean is six. Fewer than four events includes counts zero through three.

Pr⁡(T<4)=e−6∑k=036kk!\Pr(T<4)=e^{-6}\sum_{k=0}^3\frac{6^k}{k!}

Compute

Compute

Adding the first four Poisson masses gives 61 times the zero-count mass, or approximately 0.151204.

Pr⁡(T<4)=e−6(1+6+18+36)=61e−6=0.151204\Pr(T<4)=e^{-6}(1+6+18+36)=61e^{-6}=0.151204

Answer

Answer

The probability of fewer than four events is about 0.15.

0.15(B)\boxed{0.15\quad\text{(B)}}