SOA Exam P Sample Question #603. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #603 — Verified Solution

Independent solution

How to solve this Bayes' Theorem question

Answer in brief

The zero-count relationship determines the first class's Poisson mean as 3-ln(3)=1.901388. Combining the two exact-count likelihoods with prior weights 0.40 and 0.60 gives posterior probability 0.445490, so choice D is correct.

Setup

Let S denote the first class and N the second. Their prior probabilities are 0.40 and 0.60, and the second class has Poisson mean three.

\Pr(S)=0.40,\qquad \Pr(N)=0.60

\lambda_N=3

Model

Translate the zero-count comparison into an equation for the unknown mean of the first class.

e^{-\lambda_S}=3e^{-3}

\lambda_S=3-\ln 3=1.9013877113\ldots

Compute

Evaluate the two likelihoods at count two and apply Bayes' theorem.

\Pr(X=2\mid S)=e^{-\lambda_S}\frac{\lambda_S^2}{2!}=0.2699909325\ldots

\Pr(X=2\mid N)=e^{-3}\frac{3^2}{2!}=0.2240418077\ldots

\Pr(S\mid X=2)=\frac{0.40(0.2699909325)}{0.40(0.2699909325)+0.60(0.2240418077)}

\Pr(S\mid X=2)=0.4454901561\ldots

Answer

The posterior probability rounds to 0.45.

\boxed{0.45\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.

AThis is the joint probability Pr(S and X=2), approximately 0.108. It is the Bayes numerator before normalization.
BThis is the likelihood Pr(X=2 given S), approximately 0.270. It does not incorporate either prior class weight.
CThis is the prior probability of class S. Observing the count changes that probability because the two Poisson likelihoods differ.
EThis is the posterior complement, the probability of the other class after the count is observed.

Original practice · fully worked

Original variant: source allocation after a combined count

Two independent particle counters register Poisson counts with means 2 and 3 during the same interval. The combined display reports exactly four particles. Conditional on that total, calculate the probability that the first counter registered at least two particles.

A
0.1296
B
0.3456
C
0.4752
D
0.5248
E
0.8704

Variant answer in brief

Given the combined total, the first source count is binomial with four trials and success probability 2/5. Its probability of at least two is 1-0.1296-0.3456=0.5248, selecting choice D.

Setup

Let X and Y be the independent source counts and T their observed sum.

X\sim\operatorname{Poisson}(2),\qquad Y\sim\operatorname{Poisson}(3)

T=X+Y=4

Model

Independent Poisson sources split a fixed combined total in proportion to their means.

X\mid(T=4)\sim\operatorname{Binomial}\!\left(4,\frac{2}{2+3}\right)

Compute

Use the complement of zero or one particle from the first counter.

\Pr(X\ge2\mid T=4)=1-\left(\frac35\right)^4-4\left(\frac25\right)\left(\frac35\right)^3

\Pr(X\ge2\mid T=4)=1-0.1296-0.3456=0.5248

Answer

The conditional allocation probability is 0.5248.

\boxed{0.5248\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #603 — Verified Solution.” Verified 2026-07-30.

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