SOA Exam P Sample Question #633. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #633 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

Conditioning two independent Poisson counts on their sum produces a binomial allocation with success probability 2/(2+3). The probability of six allocations to the first source out of eight is 0.04128768, so the answer is C.

Setup

Let A and B be the independent counts with Poisson means 2 and 3.

A\sim\operatorname{Poisson}(2),\qquad B\sim\operatorname{Poisson}(3)

Model

Given the combined count, each event is assigned to the first source with probability proportional to its Poisson rate.

A\mid(A+B=8)\sim\operatorname{Binomial}\left(8,\frac25\right)

Compute

Evaluate the conditional binomial mass at six.

\Pr(A=6\mid A+B=8)=\binom86\left(\frac25\right)^6\left(\frac35\right)^2

\Pr(A=6\mid A+B=8)=0.04128768

Answer

The probability rounds to 0.041.

\boxed{0.041\quad\text{(C)}}

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 approximately the unconditioned joint probability of six first-source events and two second-source events; it omits division by the probability of a total of eight.
BThis is approximately the unconditional Poisson probability that the first source alone has count six.
DThis comes from perturbing the conditional source share instead of using the rate ratio 2/(2+3).
EThis substantially overweights allocations to the lower-rate source and does not follow the conditional binomial mass.

Original practice · fully worked

Original variant: wafer defect classification

A scanner records three independent classes of microscopic wafer defects. Their respective Poisson means per wafer are 1, 2, and 3. Given that the scanner finds six defects in total on a wafer, calculate the probability that the three class counts are exactly 1, 2, and 3, respectively.

A
0.002315
B
0.022309
C
0.082305
D
0.138889
E
0.312500

Variant answer in brief

Conditioned on six total defects, the three class labels have multinomial probabilities 1/6, 2/6, and 3/6. The allocation probability for counts 1, 2, and 3 is 5/36=0.1388889, so choice D is correct.

Setup

Let the three independent class counts have the stated Poisson means and condition on their combined count.

(\lambda_1,\lambda_2,\lambda_3)=(1,2,3)

\lambda_1+\lambda_2+\lambda_3=6

Model

Independent Poisson categories conditioned on their total follow a multinomial allocation with rate-proportional cell probabilities.

(K_1,K_2,K_3)\mid(K_1+K_2+K_3=6)

\sim\operatorname{Multinomial}\!\left(6;\frac16,\frac26,\frac36\right)

Compute

Evaluate the multinomial mass at the ordered class-count vector.

\Pr((K_1,K_2,K_3)=(1,2,3)\mid K_1+K_2+K_3=6)

=\frac{6!}{1!2!3!}\left(\frac16\right)\left(\frac26\right)^2\left(\frac36\right)^3

=\frac5{36}=0.1388888889\ldots

Answer

The conditional allocation probability rounds to 0.138889.

\boxed{0.138889\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #633 — Verified Solution.” Verified 2026-07-30.

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