Verified solution · Exam P

SOA Exam P Sample Question #172 Solution

This Exam P sample reference tests Conditional Expectation. The observation restricts the first count to 3, 4, or 5. Their prior-times-likelihood weights are 0.0576, 0.0864, and 0.03888; normalizing and averaging gives 495/127=3.89764, choice C.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Conditional Expectation question

Setup

Only three first-city counts can produce the observed second-city count.

J\in\{3,4,5\}\quad\text{when }K=5

Model

Multiply each binomial prior mass by its conditional observation likelihood.

w_3=\binom53(0.6)^3(0.4)^2\frac16=0.0576

w_4=\binom54(0.6)^4(0.4)\frac13=0.0864

w_5=(0.6)^5\frac12=0.03888

Compute

Normalize the three weights inside the conditional expectation.

w_3+w_4+w_5=0.18288

E[J\mid K=5]=\frac{3w_3+4w_4+5w_5}{0.18288}

E[J\mid K=5]=\frac{495}{127}=3.8976378

Answer

The conditional expected count rounds to 3.9.

\boxed{3.9\quad\text{(C)}}

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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 3.5 averages only the two lower feasible counts and discards the positive posterior weight on J=5.
BA value near 3.7 can result from conditioning the binomial prior only on J≥3. That prior tail mean is 3.608 and ignores the unequal likelihoods of producing K=5.
DThe value 4.0 is the unweighted average of 3, 4, and 5. Their posterior weights are approximately 0.315, 0.472, and 0.213, not equal.
EThe value 5.7 is impossible because the conditional support of J is {3,4,5}; an expectation cannot exceed 5.

Original practice · fully worked

Original variant: posterior mean of a hidden Poisson rate

Tickets in a maintenance archive carry source label A 60% of the time and label B otherwise. For label A the alert count follows a Poisson law with mean 1; for label B it follows a Poisson law with mean 2. One randomly chosen ticket displays exactly one alert. Find the posterior expected Poisson mean attached to its source label.

A
0.3291
B
1.0000
C
1.3291
D
1.4000
E
2.0000

Variant answer in brief

The likelihood weights for one alert are 0.6 exp(−1) and 0.8 exp(−2). The posterior probability of line B is 0.329087, so the posterior mean of the line rate is 1+0.329087=1.329087, choice C.

Setup

Compute the one-alert likelihood under each source line.

\Pr(N=1\mid A)=e^{-1}

\Pr(N=1\mid B)=2e^{-2}

Model

Combine each likelihood with its prior source probability.

w_A=0.60e^{-1}

w_B=0.40(2e^{-2})=0.80e^{-2}

Compute

Normalize the source weights and average the two possible rates.

\Pr(B\mid N=1)=\frac{0.80e^{-2}}{0.60e^{-1}+0.80e^{-2}}=0.3290869

E[\lambda\mid N=1]=1(0.6709131)+2(0.3290869)=1.3290869

Answer

The posterior expected rate is approximately 1.3291.

\boxed{1.3291\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #172 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Conditional Expectation question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: posterior mean of a hidden Poisson rate

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution