SOA Exam P Sample Question #693. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #693

Memorylessness converts the stated conditional survival into a 15-year survival probability of 0.40. Scaling the exponential survival to 50 years gives 0.047156, which rounds to choice B.

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

Answer in brief

Memorylessness converts the stated conditional survival into a 15-year survival probability of 0.40. Scaling the exponential survival to 50 years gives 0.047156, which rounds to choice B.

Setup

Let T be the exponential future lifetime and write its survival function in terms of the unknown rate.

\Pr(T\ge t)=e^{-\lambda t}

Model

The memoryless property says the conditional survival from year 10 to year 25 depends only on the additional 15 years.

\Pr(T\ge25\mid T\ge10)=\Pr(T\ge15)=e^{-15\lambda}=0.40

Compute

Express the 50-year survival as a power of the calibrated 15-year survival probability.

e^{-50\lambda}=\left(e^{-15\lambda}\right)^{50/15}

\Pr(T\ge50)=0.40^{10/3}=0.0471556032\ldots

Answer

The 50-year survival probability rounds to 0.05 at the precision of the choices.

\boxed{\Pr(T\ge50)\approx0.05\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.

AAn exponential survival probability is strictly positive at every finite time, so rounding this result all the way to zero is not justified.
CThis is approximately \(0.40^{2.5}\), which treats 50 years as 2.5 blocks of 20 years instead of using the calibrated 15-year block.
DThis is \(0.40^2\), so it accounts for only two 15-year survival blocks, or 30 years, rather than 50 years.
EThis value exceeds \(0.40^2=0.16\), even though 50 years is longer than two calibrated 15-year blocks; it therefore violates monotonicity of survival.

Original practice · fully worked

Original variant: underground moisture sensor

The operating lifetime M of an underground moisture sensor is exponential. Given that a sensor has operated for at least 6 months, the probability that it operates for at least 18 months is 0.55. Calculate the probability that a new sensor operates for at least 42 months.

A
0.0503
B
0.0915
C
0.1234
D
0.1664
E
0.5500

Variant answer in brief

The conditional statement calibrates a 12-month survival probability of 0.55. Forty-two months is 3.5 such intervals, so the survival probability is 0.123387 and choice C.

Setup

Represent the exponential survival function by a monthly rate theta.

\Pr(M\ge t)=e^{-\theta t}

Model

Memorylessness removes the six months already survived and leaves a 12-month future interval.

\Pr(M\ge18\mid M\ge6)=\Pr(M\ge12)=e^{-12\theta}=0.55

Compute

Write the 42-month survival in units of the calibrated 12-month interval.

\Pr(M\ge42)=\left(e^{-12\theta}\right)^{42/12}

0.55^{3.5}=0.1233870023\ldots

Answer

A new sensor reaches 42 months with probability about 0.1234.

\boxed{0.1234\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #693 — Verified Solution.” Verified 2026-07-30.

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