Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let U be the daily count in the marked category. The total count has mean 5, and an item enters U with probability 0.25.

λ=5\lambda=5
q=0.25q=0.25

Model

Model

Poisson thinning gives another Poisson variable whose mean is the original mean times the marking probability.

U∼Poisson⁡(λq)U\sim\operatorname{Poisson}(\lambda q)
λU=5(0.25)=1.25\lambda_U=5(0.25)=1.25

Compute

Compute

Evaluate the probability mass at zero for the thinned count.

Pr⁡(U=0)=e−1.251.2500!=e−1.25\Pr(U=0)=e^{-1.25}\frac{1.25^0}{0!}=e^{-1.25}
e−1.25=0.2865047969e^{-1.25}=0.2865047969

Answer

Answer

Rounding to the listed precision gives 0.287.

0.287(E)\boxed{0.287\quad\text{(E)}}