Independent solution

How to solve this Conditional Distributions question

Answer in brief

Conditional on a first-task duration h, the second duration is uniform on the integers from 1 through h-1 and therefore has mean h/2. Averaging over the three equally likely h values gives 11/6, choice A.

Setup

Setup

Let H denote the first duration and Y the second duration. The three possible values of H have equal probability.

Pr⁡(H=2)=Pr⁡(H=4)=Pr⁡(H=5)=13\Pr(H=2)=\Pr(H=4)=\Pr(H=5)=\frac13

Model

Model

For a fixed value h, Y is discrete uniform on 1 through h-1. Its conditional mean is the midpoint of the first and last values.

E⁡[Y∣H=h]=1+(h−1)2=h2\operatorname{E}[Y\mid H=h]=\frac{1+(h-1)}{2}=\frac h2

Compute

Compute

Apply the law of total expectation to the three conditional means.

E⁡[Y]=13(22+42+52)\operatorname{E}[Y]=\frac13\left(\frac22+\frac42+\frac52\right)
E⁡[Y]=1+2+2.53=116\operatorname{E}[Y]=\frac{1+2+2.5}{3}=\frac{11}{6}

Answer

Answer

The expected second duration is one and five-sixths hours.

E⁡[Y]=116=156(A)\boxed{\operatorname{E}[Y]=\frac{11}{6}=1\frac56\quad\text{(A)}}