Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let N be the count and write conditional expectation as a truncated first moment divided by the conditioning probability.

N∼Poisson⁡(1)N\sim\operatorname{Poisson}(1)
E⁡[N∣N≥2]=E⁡[N1{N≥2}]Pr⁡(N≥2)\operatorname{E}[N\mid N\ge2]=\frac{\operatorname{E}[N\mathbf{1}_{\{N\ge2\}}]}{\Pr(N\ge2)}

Model

Model

Remove the count-one contribution from the unconditional first moment; count zero contributes nothing.

E⁡[N1{N≥2}]=E⁡[N]−Pr⁡(N=1)=1−e−1\operatorname{E}[N\mathbf{1}_{\{N\ge2\}}]=\operatorname{E}[N]-\Pr(N=1)=1-e^{-1}
Pr⁡(N≥2)=1−Pr⁡(N=0)−Pr⁡(N=1)=1−2e−1\Pr(N\ge2)=1-\Pr(N=0)-\Pr(N=1)=1-2e^{-1}

Compute

Compute

Divide the truncated first moment by the remaining probability mass.

E⁡[N∣N≥2]=1−e−11−2e−1=2.3922112\operatorname{E}[N\mid N\ge2]=\frac{1-e^{-1}}{1-2e^{-1}}=2.3922112

Answer

Answer

The conditional expected count is approximately 2.39.

2.39(B)\boxed{2.39\quad\text{(B)}}