SOA Exam P Sample Question #699. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #699 — Verified Solution

Independent solution

How to solve this Bayes' Theorem question

Answer in brief

Bayes' theorem combines the 20% prior class share with the two class-specific Poisson likelihoods for a count of two. The posterior probability is 0.170298, which matches choice B.

Setup

Let A denote membership in the lower-rate class and let N be the observed annual count. Record the class proportions and Poisson means.

\Pr(A)=0.20,\qquad \Pr(A^c)=0.80

N\mid A\sim\operatorname{Poisson}(1),\qquad N\mid A^c\sim\operatorname{Poisson}(3)

Model

Evaluate the likelihood of exactly two events in each class and place those likelihoods into Bayes' formula.

\Pr(N=2\mid A)=\frac{e^{-1}}{2},\qquad \Pr(N=2\mid A^c)=\frac{3^2e^{-3}}{2}

\Pr(A\mid N=2)=\frac{0.20(e^{-1}/2)}{0.20(e^{-1}/2)+0.80(9e^{-3}/2)}

Compute

Calculate the weighted likelihood in the numerator and the total mixture likelihood in the denominator.

\Pr(A,N=2)=0.0367879441\ldots

\Pr(N=2)=0.0367879441\ldots+0.1792334461\ldots=0.2160213902\ldots

\Pr(A\mid N=2)=0.1702976917\ldots

Answer

The requested posterior class probability rounds to 0.170.

\boxed{\Pr(A\mid N=2)\approx 0.170\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 0.20(e^{-1}/2)=0.036788 is the joint numerator P(A and N=2), not the posterior after division by the mixture likelihood.
CThe value e^{-1}/2=0.183940 is the within-class likelihood of two events and omits both the prior and Bayes normalization.
DThe value 0.200 is the original class proportion; it ignores the information supplied by the observed count.
EThe value (e^{-1}/2)/(e^{-1}/2+9e^{-3}/2)=0.450853 treats the two classes as equally likely before observing the count.

Original practice · fully worked

Original variant: sensor warning classification

A warehouse uses two sensor models. Model R makes up 30% of installed sensors and produces a Poisson number of warning signals per week with mean 0.5. Model S makes up the remainder and has mean 2. Given that a randomly selected sensor produced exactly one warning last week, calculate the probability that it is model R.

A
0.091
B
0.300
C
0.303
D
0.324
E
0.528

Variant answer in brief

Weighting the two one-warning likelihoods by the 30% and 70% model shares gives a posterior model-R probability of 0.324407, so choice D is correct.

Setup

Let W be the weekly warning count and write the prior model shares and conditional means.

\Pr(R)=0.30,\qquad \Pr(S)=0.70

W\mid R\sim\operatorname{Poisson}(0.5),\qquad W\mid S\sim\operatorname{Poisson}(2)

Model

Use the one-warning mass under each model as the likelihood in Bayes' formula.

\Pr(W=1\mid R)=0.5e^{-0.5},\qquad \Pr(W=1\mid S)=2e^{-2}

\Pr(R\mid W=1)=\frac{0.30(0.5e^{-0.5})}{0.30(0.5e^{-0.5})+0.70(2e^{-2})}

Compute

Evaluate the weighted likelihoods and normalize.

\Pr(R,W=1)=0.0909795990\ldots

\Pr(W=1)=0.0909795990\ldots+0.1894693965\ldots=0.2804489955\ldots

\Pr(R\mid W=1)=0.3244069347\ldots

Answer

The posterior probability is approximately 0.324.

\boxed{\Pr(R\mid W=1)\approx 0.324\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #699 — Verified Solution.” Verified 2026-07-30.

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