Worked solution · Exam P

SOA Exam P Sample Question #525 Solution

This Exam P sample reference tests Joint Distributions. After conditioning on the news-viewing group, the count has probabilities 1/14, 3/7, and 1/2 at 0, 1, and 2. Its variance is 19/49=0.387755, so choice B is correct.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Joint Distributions question

Setup

Let K count the two entertainment indicators. Restrict the table to the conditioning category, whose total mass is 0.70, and aggregate cells that have the same value of K.

p_0=\frac{0.05}{0.70}=\frac{1}{14},\qquad p_1=\frac{0.08+0.22}{0.70}=\frac{3}{7}

p_2=\frac{0.35}{0.70}=\frac{1}{2}

Model

Compute the first two conditional moments of the count.

E[K\mid N]=p_1+2p_2

E[K^2\mid N]=p_1+4p_2

Compute

Substitute the three conditional masses and subtract the square of the mean.

E[K\mid N]=\frac{10}{7},\qquad E[K^2\mid N]=\frac{17}{7}

\operatorname{Var}(K\mid N)=\frac{17}{7}-\left(\frac{10}{7}\right)^2=\frac{19}{49}=0.387755

Answer

The requested conditional variance rounds to 0.388.

\boxed{0.388\quad\text{(B)}}

Test the method on unseen problems

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30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis turns K into an even-versus-odd indicator. Since P(K is even | N)=4/7, that indicator has variance 12/49=0.2449, whose nearest option is 0.242.
CThis rounds the two conditional category rates to 0.6 and 0.8 and ignores their covariance, giving 0.6(0.4)+0.8(0.2)=0.400.
DThis ignores the conditioning. The unconditional count has first moment 1.32 and second moment 2.22, so its variance is 2.22-1.32²=0.4776.
EThis reports the conditional probability that K is even, (0.35+0.05)/0.70=4/7=0.5714, instead of the variance of K.

Original practice · fully worked

Original variant: diagnostic-flag count for one supplier

A quality lab records K, the number of three diagnostic flags triggered by a component. The joint masses for supplier Q are P(Q,K=0)=0.08, P(Q,K=1)=0.24, P(Q,K=2)=0.32, and P(Q,K=3)=0.16. Calculate Var(K | Q).

A
0.09
B
0.81
C
1.11
D
2.00
E
3.70

Answer and calculation

The Q masses total 0.80, giving conditional probabilities 0.10, 0.30, 0.40, and 0.20. Thus E[K]=1.70, E[K²]=3.70, and Var(K)=0.81, so choice B is correct.

Setup

Normalize the four joint masses by their total for supplier Q.

P(Q)=0.08+0.24+0.32+0.16=0.80

(p_0,p_1,p_2,p_3)=(0.10,0.30,0.40,0.20)

Model

Use the conditional first and second moments of the flag count.

E[K\mid Q]=\sum_{k=0}^{3}kp_k,\qquad E[K^2\mid Q]=\sum_{k=0}^{3}k^2p_k

Compute

Evaluate both moments and then center the second moment.

E[K\mid Q]=1.70,\qquad E[K^2\mid Q]=3.70

\operatorname{Var}(K\mid Q)=3.70-(1.70)^2=0.81

Answer

The conditional flag-count variance is 0.81.

\boxed{0.81\quad\text{(B)}}

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How to solve this Joint Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: diagnostic-flag count for one supplier

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables