SOA Exam P Sample Question #638. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #638 — Verified Solution

Independent solution

How to solve this Conditional Distributions question

Answer in brief

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.

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)}}

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

Variant answer in brief

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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ActuaryProof. “SOA Exam P Sample Question #638 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.