SOA Exam P Sample Question #621. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #621 — Verified Solution

Independent solution

How to solve this Exponential Distribution question

Answer in brief

The exponential memoryless property makes the elapsed event-free period irrelevant to the residual wait. Subtracting the survival probability beyond 25 days from the survival probability beyond 5 days gives 0.527656, which rounds to 0.53 and selects choice D.

Setup

Let W denote the additional waiting time from the present. The original exponential mean determines its rate.

\lambda=\frac{1}{15}

Model

By memorylessness, the residual wait is exponential with the same mean despite the elapsed quiet period.

\Pr(W>t)=e^{-t/15}

Compute

The probability of landing between the two residual-time bounds is the difference of their survival probabilities.

\Pr(5<W<25)=e^{-5/15}-e^{-25/15}

\Pr(5<W<25)=0.5276557077\ldots

Answer

The requested probability rounds to 0.53.

\boxed{0.53\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AComputing the unconditional probability between absolute times 15 and 35 gives about 0.271 and fails to divide by the survival probability through the elapsed 10 days.
BThis is the exponential survival probability beyond one mean, e^-1, and does not represent the bounded residual interval.
CThis is the complement of the desired interval probability, covering a residual wait at most 5 or at least 25 days.
EThis is approximately the probability that a fresh exponential wait is at most its mean, which replaces the two given residual bounds by a single 15-day cutoff.

Original practice · fully worked

Original variant: delayed calibration race

At a remote observatory, the time until a calibration routine completes is exponential with mean 8 minutes. An independent safety watchdog stops the routine after an exponential time with mean 12 minutes. Both clocks start together. Calculate the probability that calibration finishes first and finishes more than 3 minutes after the start.

A
0.214
B
0.321
C
0.535
D
0.600
E
0.687

Variant answer in brief

The two clocks race at combined rate 5/24. Integrating the calibration completion density after minute three while the watchdog survives gives (3/5)e^(-5/8)=0.321157, so choice B is correct.

Setup

Let C be the calibration completion time and W the independent watchdog time. Convert their means to rates.

C\sim\operatorname{Exp}(1/8),\qquad W\sim\operatorname{Exp}(1/12)

Model

For calibration to finish first at time t, its density contributes while the watchdog must survive through t.

\Pr(C<W,\ C>3)=\int_3^\infty \frac18e^{-t/8}e^{-t/12}\,dt

Compute

Combine the exponential rates and evaluate the tail integral.

\frac18+\frac1{12}=\frac5{24}

\Pr(C<W,\ C>3)=\frac{1/8}{5/24}e^{-(5/24)3}

\Pr(C<W,\ C>3)=\frac35e^{-5/8}=0.3211568571\ldots

Answer

The delayed calibration-win probability rounds to 0.321.

\boxed{0.321\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #621 — Verified Solution.” Verified 2026-07-30.

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