SOA Exam P Sample Question #617. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #617 — Verified Solution

Independent solution

How to solve this Conditional Distributions question

Answer in brief

Normalize the three joint masses in the specified column to obtain conditional probabilities 0.56, 0.32, and 0.12. The conditional second moment is 0.80 and the conditional mean is 0.56, so the variance is 0.4864 and choice D is correct.

Setup

The conditioning column has total probability 0.25. Divide each mass in that column by this total before calculating any moment.

\Pr(F=1)=0.14+0.08+0.03=0.25

\Pr(T=0,1,2\mid F=1)=(0.56,0.32,0.12)

Model

Use the first two raw moments of the resulting conditional distribution.

\operatorname{E}[T\mid F=1]=0(0.56)+1(0.32)+2(0.12)=0.56

\operatorname{E}[T^2\mid F=1]=0^2(0.56)+1^2(0.32)+2^2(0.12)=0.80

Compute

Subtract the square of the conditional mean from the conditional second moment.

\operatorname{Var}(T\mid F=1)=0.80-(0.56)^2

\operatorname{Var}(T\mid F=1)=0.4864

Answer

The variance rounds to 0.49.

\boxed{0.49\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

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

AUsing the unnormalized column masses as though they formed a probability distribution gives 0.20-(0.14)^2=0.1804, which rounds to 0.18.
BSubtracting the mean itself rather than its square from the second moment gives 0.80-0.56=0.24.
CIgnoring the conditioning column and using the marginal distribution of T gives an unconditional variance of 0.45.
EThis is the conditional second moment 0.80; it omits subtraction of the squared conditional mean.

Original practice · fully worked

Original variant: satellite diagnostic update

A satellite records zero, one, or two anomaly pings in a cycle with prior probabilities 0.50, 0.30, and 0.20. A diagnostic lamp activates with probability 0.10 after zero pings, 0.40 after one ping, and 0.80 after two pings. Given that the lamp activated, find the variance of the ping count.

A
0.330
B
0.525
C
0.610
D
1.333
E
2.303

Variant answer in brief

Bayes' rule turns the three prior-by-likelihood products into posterior masses 5/33, 12/33, and 16/33. Their variance is 52/99, or 0.525253, so choice B is correct.

Setup

Let N be the ping count and A the event that the diagnostic lamp activates. Record the prior probabilities and the count-specific activation likelihoods.

\bigl(\Pr(N=0),\Pr(N=1),\Pr(N=2)\bigr)=(0.50,0.30,0.20)

\bigl(\Pr(A\mid N=0),\Pr(A\mid N=1),\Pr(A\mid N=2)\bigr)=(0.10,0.40,0.80)

Model

Multiply each prior by its activation likelihood. Their sum is the evidence used to normalize the posterior distribution.

(0.50)(0.10),(0.30)(0.40),(0.20)(0.80)=(0.05,0.12,0.16)

\Pr(A)=0.05+0.12+0.16=0.33

\bigl(\Pr(N=0\mid A),\Pr(N=1\mid A),\Pr(N=2\mid A)\bigr)=\left(\frac5{33},\frac{12}{33},\frac{16}{33}\right)

Compute

Calculate the first two posterior moments and apply the variance identity.

\operatorname{E}[N\mid A]=\frac{12+2(16)}{33}=\frac43

\operatorname{E}[N^2\mid A]=\frac{12+4(16)}{33}=\frac{76}{33}

\operatorname{Var}(N\mid A)=\frac{76}{33}-\left(\frac43\right)^2=\frac{52}{99}=0.525252\ldots

Answer

The posterior variance rounds to 0.525.

\boxed{0.525\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #617 — Verified Solution.” Verified 2026-07-30.

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