Verified solution · Exam P

SOA Exam P Sample Question #187 Solution

This Exam P sample reference tests Mixture Distributions. The within-policy matching probabilities are 0.30 for type A and 0.25 for type B, giving an overall matching probability of 0.285. A Binomial(4,0.285) count equals one with probability 0.41670, choice D.

Independent analysis by ActuaryProof · reviewed August 4, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Mixture Distributions question

Setup

For each policy type, add the squared claim-count probabilities because the two cars are independent.

p_A=0.40^2+0.30^2+0.20^2+0.10^2=0.30

p_B=4(0.25^2)=0.25

Model

Average the matching probability over the policy-type mixture.

p=0.70(0.30)+0.30(0.25)=0.285

Compute

Treat the four independently selected policies as Bernoulli trials for the matching event.

K\sim\operatorname{Binomial}(4,0.285)

\Pr(K=1)=\binom41(0.285)(0.715)^3=0.4166995

Answer

The probability rounds to 0.417.

\boxed{0.417\quad\text{(D)}}

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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 0.104 is approximately 0.285(0.715)³, the probability for one specified policy to match while the other three do not. It omits the four choices of the matching policy.
BThe value 0.250 is the matching probability for a type-B policy only. It does not average over policy types or consider four selections.
CThe value 0.285 is the matching probability for one randomly selected policy. It is the binomial trial probability, not the exactly-one count probability.
EThe value 0.739 is 1-(0.715)⁴, the probability that at least one of the four policies has matching counts. It also includes two, three, and four matches.

Original practice · fully worked

Original variant: infer a latent mode from matching codes

A device uses calibration mode A with probability 0.70 and mode B otherwise. Conditional on mode A, two independent output codes each take values 1, 2, and 3 with respective probabilities 60%, 25%, and 15%. Under mode B, each code is uniform on those three values. Given a matching pair of output codes, determine the chance that the device used mode A.

A
0.3333
B
0.4115
C
0.4450
D
0.7000
E
0.7570

Variant answer in brief

The matching probabilities are 0.445 in mode A and 1/3 in mode B. Bayes' rule gives 0.70(0.445)/[0.70(0.445)+0.30(1/3)]=0.7570, choice E.

Setup

Compute the collision probability under each calibration mode.

\Pr(M\mid A)=0.60^2+0.25^2+0.15^2=0.445

\Pr(M\mid B)=3\left(\frac13\right)^2=\frac13

Model

Combine the two modes to obtain the marginal matching probability.

\Pr(M)=0.70(0.445)+0.30\left(\frac13\right)=0.4115

Compute

Apply Bayes' rule to the observed match.

\Pr(A\mid M)=\frac{0.70(0.445)}{0.4115}

\Pr(A\mid M)=\frac{0.3115}{0.4115}=0.7569866

Answer

Given matching codes, mode A has posterior probability about 0.7570.

\boxed{0.7570\quad\text{(E)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #187 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Mixture Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: infer a latent mode from matching codes

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution