Verified solution · Exam P

SOA Exam P Sample Question #605 Solution

This Exam P sample reference tests Bayes' Theorem. The three-year failure probabilities are 1-exp(-3/2) for type A and 1-exp(-3/4) for type B. Bayes' theorem with prior weights 0.10 and 0.90 gives posterior type-A probability 0.140595, so choice B is correct.

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 Bayes' Theorem question

Setup

Let A and B identify the two product types, and let F be failure before year three.

\Pr(A)=0.10,\qquad \Pr(B)=0.90

F=\{T<3\}

Model

Use each exponential survival function to calculate the likelihood of the observed early failure.

\Pr(F\mid A)=1-e^{-3/2}=0.7768698399\ldots

\Pr(F\mid B)=1-e^{-3/4}=0.5276334473\ldots

Compute

Weight the two likelihoods by their prior probabilities and normalize.

\Pr(A\mid F)=\frac{0.10(1-e^{-3/2})}{0.10(1-e^{-3/2})+0.90(1-e^{-3/4})}

\Pr(A\mid F)=0.1405953989\ldots

Answer

The posterior probability rounds to 0.14.

\boxed{0.14\quad\text{(B)}}

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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.

AThis is the joint probability of type A and early failure, about 0.0777. It is the Bayes numerator before division by the total early-failure probability.
CThis is the probability that a type-B item survives three years, exp(-3/4). It reverses the observed event and omits Bayes' theorem.
DThis is close to the overall early-failure probability, the denominator that normalizes the two weighted likelihoods. It is not a type posterior.
EThis is the likelihood of early failure for type A, 1-exp(-3/2), without accounting for the small prior share of that type.

Original practice · fully worked

Original variant: one module failure by inspection

A research rack contains four independent modules whose lifetimes are exponential with mean 5 years. At an inspection two years after installation, calculate the probability that exactly one module has failed.

A
0.2019
B
0.3297
C
0.3972
D
0.7981
E
1.3187

Variant answer in brief

Each module fails by year two with probability 1-exp(-0.4). The number failed is binomial with four trials, giving exactly-one probability 4(1-exp(-0.4))exp(-1.2)=0.397191 and choice C.

Setup

Compute the failure and survival probabilities for one module at the inspection time.

p=\Pr(T\le2)=1-e^{-2/5}=1-e^{-0.4}

1-p=e^{-0.4}

Model

Independence makes the number of failed modules at year two binomial.

K\sim\operatorname{Binomial}\!\left(4,1-e^{-0.4}\right)

Compute

Choose which one module failed and require the other three to survive.

\Pr(K=1)=\binom41(1-e^{-0.4})(e^{-0.4})^3

\Pr(K=1)=4(1-e^{-0.4})e^{-1.2}=0.3971907757\ldots

Answer

The exactly-one probability rounds to 0.3972.

\boxed{0.3972\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #605 — Verified Solution.” Verified 2026-07-30.

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

How to solve this Bayes' Theorem question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: one module failure by inspection

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution