Worked solution · Exam P

SOA Exam P Sample Question #631 Solution

This Exam P sample reference tests Conditional Distributions. This is a conditional-variance calculation for a binary random variable. Within the specified row, the conditional probability of one is 4/5, so the Bernoulli variance is (4/5)(1/5)=4/25 and choice A is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 Conditional Distributions question

Setup

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

\Pr(X=1)=\frac18+\frac12=\frac58

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\mid X=1)=\frac{1/2}{5/8}=\frac45

Compute

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

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

\operatorname{Var}(Y\mid X=1)=\frac45\left(1-\frac45\right)=\frac4{25}=0.16

Answer

The conditional variance is four twenty-fifths.

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

Test the method on unseen problems

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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.

BThe fraction 15/64 is (5/8)(3/8), the variance of the indicator that X equals one. It uses the conditioning-row probability as a Bernoulli parameter instead of forming Y's conditional distribution.
CUsing the unnormalized joint mass 1/2 as though it were the conditional success probability gives (1/2)(1/2)=1/4.
DThe value 3/4 is the complement of the separate joint cell with mass 1/4; taking that complement does not condition Y on X equal to one.
EThe fraction 4/5 is the conditional probability and conditional mean of Y, not its conditional variance.

Original practice · fully worked

Original variant: warning-conditioned calibration spread

A sensor line operates in fast mode with probability 0.60 and precision mode with probability 0.40. A warning appears with probability 0.20 in fast mode and 0.50 in precision mode. Given the mode, the warning is independent of calibration adjustment Y. In fast mode, Y has mean 0 and variance 1; in precision mode, it has mean 3 and variance 2. Given that a warning appeared, calculate Var(Y).

A
0.3200
B
1.6250
C
1.8750
D
2.1094
E
3.7344

Answer and calculation

Bayes' rule gives posterior mode weights 3/8 for fast and 5/8 for precision after a warning. The conditional variance is the 1.625 within-mode component plus the 2.109375 between-mode component, totaling 3.734375 and selecting choice E.

Setup

Let W denote a warning. Compute its probability and update the operating-mode weights using Bayes' rule.

\Pr(W)=0.60(0.20)+0.40(0.50)=0.32

\Pr(F\mid W)=\frac{0.60(0.20)}{0.32}=\frac38,\qquad \Pr(P\mid W)=\frac58

Model

Apply the conditional law of total variance across the two posterior modes.

\operatorname{Var}(Y\mid W)=\operatorname{E}[\operatorname{Var}(Y\mid M,W)\mid W]+\operatorname{Var}(\operatorname{E}[Y\mid M,W]\mid W)

Compute

Conditional independence allows the supplied mode-specific moments to be used after observing the warning.

\operatorname{E}[\operatorname{Var}(Y\mid M,W)\mid W]=\frac38(1)+\frac58(2)=1.625

\operatorname{E}[Y\mid W]=\frac38(0)+\frac58(3)=1.875

\operatorname{Var}(\operatorname{E}[Y\mid M,W]\mid W)=\frac38(0-1.875)^2+\frac58(3-1.875)^2=2.109375

\operatorname{Var}(Y\mid W)=1.625+2.109375=3.734375

Answer

The calibration adjustment has conditional variance approximately 3.7344 after a warning.

\boxed{3.7344\quad\text{(E)}}

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How to solve this Conditional Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: warning-conditioned calibration spread

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables