Independent solution

How to solve this Conditional Distributions question

Setup

Setup

Conditioning the joint count law on the April count leaves a May count supported on positive integers.

P(N2=n∣N1=2)=e−2(1−e−2)n−1,n≥1P(N_2=n\mid N_1=2)=e^{-2}(1-e^{-2})^{n-1},\qquad n\ge1

Model

Model

The remaining conditional mass is geometric on 1,2,... with success probability exp(-2).

N2∣N1=2∼Geom⁡(p=e−2)N_2\mid N_1=2\sim\operatorname{Geom}(p=e^{-2})

Compute

Compute

For a trials-until-success geometric variable, the mean is the reciprocal of its success probability, giving exp(2).

E[N2∣N1=2]=1e−2=e2E[N_2\mid N_1=2]=\frac1{e^{-2}}=e^2

Answer

Answer

Therefore the conditional expected May count is exp(2), corresponding to choice E.

e2(E)\boxed{e^2\quad\text{(E)}}