Worked solution · Exam P

SOA Exam P Sample Question #621 Solution

This Exam P sample reference tests Exponential Distribution. 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.

Independent analysis by ActuaryProof · reviewed July 30, 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

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)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

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, exp(−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

Answer and calculation

The two clocks race at combined rate 5/24. Integrating the calibration completion density after minute three while the watchdog survives gives (3/5) exp(−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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: delayed calibration race

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables