Worked solution · Exam P

SOA Exam P Sample Question #638 Solution

This Exam P sample reference tests Conditional Distributions. This is a conditional-moment calculation from one row of a joint probability table. Normalizing that row gives a conditional mean of 86/33 and a second moment of 274/33, so the variance is 1646/1089, or 1.511478, and choice B 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

Condition on the stated value of the first variable. The corresponding five joint masses have total probability 0.33.

\Pr(S=2)=0.10+0.03+0.07+0.05+0.08=0.33

\Pr(B=b\mid S=2)=\frac{p_{2,b}}{0.33},\qquad b=1,\ldots,5

Model

Compute the first two conditional raw moments and then use the variance identity.

\operatorname{Var}(B\mid S=2)=\operatorname{E}[B^2\mid S=2]-\operatorname{E}[B\mid S=2]^2

Compute

Weight the five possible values by the normalized row probabilities.

\operatorname{E}[B\mid S=2]=\frac{1(0.07)+2(0.10)+3(0.08)+4(0.05)+5(0.03)}{0.33}=\frac{86}{33}

\operatorname{E}[B^2\mid S=2]=\frac{1^2(0.07)+2^2(0.10)+3^2(0.08)+4^2(0.05)+5^2(0.03)}{0.33}=\frac{274}{33}

\operatorname{Var}(B\mid S=2)=\frac{274}{33}-\left(\frac{86}{33}\right)^2=\frac{1646}{1089}=1.5114784206\ldots

Answer

The conditional variance rounds to 1.51.

\boxed{1.51\quad\text{(B)}}

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.

AThe value 1.23 is approximately the conditional standard deviation, obtained by taking the square root of the required variance.
CUsing the joint row entries without dividing by their row total gives 2.74-0.86 squared, approximately 2.00; those entries do not form a conditional distribution until normalized.
DDividing the first-moment numerator 0.86 by an incorrectly added row total of 0.35 gives about 2.46, and that quantity would be a mean rather than a variance.
EA value near 6.06 is a squared location estimate, not a variance computed as the second moment minus the square of the mean; it also exceeds the correct moment calculation by a wide margin.

Original practice · fully worked

Original variant: night-mode inspection flags

An automated panel inspector records its operating mode M and the number D of flags raised. In the night-mode row of the joint probability table, the entries for D=0, 1, 2, and 3 are 0.03, 0.09, 0.12, and 0.06; the remaining joint mass belongs to other modes. Given that the inspector is in night mode, calculate the variance of D.

A
0.300
B
0.850
C
0.810
D
1.700
E
3.700

Answer and calculation

The night-mode row has mass 0.30. After normalization, the conditional mean is 1.70 and the conditional second moment is 3.70, so the variance is 3.70 minus 1.70 squared, or 0.81; this selects choice C.

Setup

First find the probability of the conditioning mode from its joint-table row.

\Pr(M=\text{night})=0.03+0.09+0.12+0.06=0.30

Model

Divide each row entry by 0.30 to obtain the conditional distribution of the flag count.

\Pr(D=d\mid M=\text{night})=\frac{(0.03,0.09,0.12,0.06)_d}{0.30},\qquad d=0,1,2,3

Compute

Calculate both conditional moments before applying the variance identity.

\operatorname{E}[D\mid M=\text{night}]=\frac{0(0.03)+1(0.09)+2(0.12)+3(0.06)}{0.30}=1.70

\operatorname{E}[D^2\mid M=\text{night}]=\frac{0^2(0.03)+1^2(0.09)+2^2(0.12)+3^2(0.06)}{0.30}=3.70

\operatorname{Var}(D\mid M=\text{night})=3.70-1.70^2=0.81

Answer

The conditional variance of the number of flags is 0.81.

\boxed{0.810\quad\text{(C)}}

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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: night-mode inspection flags

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables