Verified solution · Exam P

SOA Exam P Sample Question #520 Solution

This Exam P sample reference tests Conditional Distributions. This is a conditional Bernoulli variance obtained by normalizing the relevant slice of a discrete joint distribution. Within Y=1, the probability that W=1 is 3/8, so the variance is 15/64 and choice B is correct.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Conditional Distributions question

Setup

After fixing Y at 1, sum the proportional weights over the two possible values of X for each value of W.

a_0=(1+1)+(2+1)=5

a_1=(1+1-1)+(2+1-1)=3

Model

Normalize the two weights. The conditional distribution of W is Bernoulli because W can only be zero or one.

p=\Pr(W=1\mid Y=1)=\frac{a_1}{a_0+a_1}=\frac{3}{8}

\operatorname{Var}(W\mid Y=1)=p(1-p)

Compute

Insert the conditional success probability into the Bernoulli variance formula.

p(1-p)=\frac{3}{8}\left(1-\frac{3}{8}\right)

\frac{3}{8}\frac{5}{8}=\frac{15}{64}

Answer

The conditional variance is 15/64.

\boxed{\frac{15}{64}\quad\text{(B)}}

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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 uses the joint probability P(W=1,Y=1)=3/20 as though it were conditional; the resulting Bernoulli calculation (3/20)(17/20)=51/400 omits normalization by P(Y=1).
CThis ignores the condition on Y. Across all cells P(W=1)=8/20=2/5, whose unconditional variance is (2/5)(3/5)=6/25.
DThis reports the conditional mean P(W=1|Y=1)=3/8. For a Bernoulli variable, that mean must still be multiplied by 1-p to obtain the variance.
EThis reports the unconditional mean P(W=1)=8/20=2/5 and therefore omits both the conditioning and the variance calculation.

Original practice · fully worked

Original variant: variance of a conditionally scored audit flag

A quality audit records binary indicators U and V. Their joint probabilities are P(U=0,V=0)=0.18, P(U=0,V=1)=0.12, P(U=1,V=0)=0.28, and P(U=1,V=1)=0.42. The audit score is S=3U-1. Calculate Var(S | V=1).

A
14/81
B
7/9
C
4/3
D
14/9
E
189/100

Variant answer in brief

Given V=1, the probability that U=1 is 0.42/(0.12+0.42)=7/9. Scaling a Bernoulli variable by 3 multiplies its variance by 9, so Var(S|V=1)=14/9 and choice D is correct.

Setup

Normalize the V=1 column of the joint table.

\Pr(U=1\mid V=1)=\frac{0.42}{0.12+0.42}=\frac{7}{9}

\Pr(U=0\mid V=1)=\frac{2}{9}

Model

A constant shift has no effect on variance, while multiplication by 3 contributes a factor of 9.

\operatorname{Var}(S\mid V=1)=\operatorname{Var}(3U-1\mid V=1)=9\operatorname{Var}(U\mid V=1)

Compute

Use the conditional Bernoulli variance and apply the scaling factor.

\operatorname{Var}(U\mid V=1)=\frac{7}{9}\frac{2}{9}=\frac{14}{81}

\operatorname{Var}(S\mid V=1)=9\frac{14}{81}=\frac{14}{9}

Answer

The conditional score variance is 14/9.

\boxed{\frac{14}{9}\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #520 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Conditional Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: variance of a conditionally scored audit flag

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution