Independent solution

How to solve this Conditional Distributions question

Setup

Setup

Add the two joint masses in the row selected by the conditioning event.

Pr⁡(X=1)=18+12=58\Pr(X=1)=\frac18+\frac12=\frac58

Model

Model

After conditioning, Y is Bernoulli. Its success probability is the joint mass for Y equal to one divided by the selected row total.

p=Pr⁡(Y=1∣X=1)=1/25/8=45p=\Pr(Y=1\mid X=1)=\frac{1/2}{5/8}=\frac45

Compute

Compute

Apply the Bernoulli variance formula with conditional success probability four-fifths.

Var⁡(Y∣X=1)=p(1−p)\operatorname{Var}(Y\mid X=1)=p(1-p)
Var⁡(Y∣X=1)=45(1−45)=425=0.16\operatorname{Var}(Y\mid X=1)=\frac45\left(1-\frac45\right)=\frac4{25}=0.16

Answer

Answer

The conditional variance is four twenty-fifths.

425(A)\boxed{\frac4{25}\quad\text{(A)}}