Verified solution · Exam P

SOA Exam P Sample Question #381 Solution

This Exam P sample reference tests Exponential Distribution. This problem asks for an exponential lifetime probability after survival to an earlier age. Memorylessness leaves a fifteen-year residual window, whose failure probability is 0.608394 and choice C.

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 Exponential Distribution question

Setup

Convert the supplied exponential rate to its survival function.

\lambda=0.0625=\frac1{16}

S(x)=e^{-x/16}

Model

Conditioning on survival to year five turns the upper endpoint into fifteen additional years.

\Pr(X<20\mid X\ge5)=1-\Pr(X\ge20\mid X\ge5)

\Pr(X\ge20\mid X\ge5)=\frac{S(20)}{S(5)}=e^{-15/16}

Compute

Evaluate the residual lower-tail probability.

1-e^{-15/16}=0.6083943733\ldots

Answer

The conditional probability rounds to 0.608.

\boxed{0.608\quad\text{(C)}}

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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 computes only the unconditional interval probability S(5)-S(20)=0.445111 and omits division by the conditioning probability S(5).
BThis incorrectly treats the nested events as independent and multiplies S(5)F(20)=0.522004, which rounds to 0.522.
DThis divides by the survival-to-five probability twice: [S(5)-S(20)]/S(5)²=0.831577, producing 0.832.
EThis divides F(20) by S(5) without removing failures before year five from the numerator, giving 0.975232.

Original practice · fully worked

Original variant: conditional failure under an increasing hazard

A filtration cartridge has survival function S(t)=exp[-(t/8)²] for t≥0, where t is measured in hours. A cartridge is still operating at hour 4. Calculate the conditional probability that it fails by hour 8.

A
0.2212
B
0.3679
C
0.5276
D
0.6321
E
0.7788

Variant answer in brief

Conditional survival from hour 4 through hour 8 is S(8)/S(4)=exp(-0.75). Its complement is 0.527633, so choice C.

Setup

Write the conditional survival ratio over the observed operating interval.

\Pr(T>8\mid T>4)=\frac{S(8)}{S(4)}

Model

The squared-time cumulative hazard means the elapsed four hours do not cancel as they would for an exponential lifetime.

\frac{S(8)}{S(4)}=\exp\!\left[-\left(\frac88\right)^2+\left(\frac48\right)^2\right]

\frac{S(8)}{S(4)}=e^{-3/4}

Compute

Take the complement of conditional survival to obtain failure by hour eight.

\Pr(T\le8\mid T>4)=1-e^{-3/4}=0.5276334473\ldots

Answer

The conditional failure probability rounds to 0.5276.

\boxed{0.5276\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #381 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: conditional failure under an increasing hazard

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution