Worked solution · Exam P

SOA Exam P Sample Question #497 Solution

This Exam P sample reference tests Exponential Distribution. Each loss has exponential rate 1/6. The minimum of three independent exponential variables has rate 3(1/6)=1/2 and therefore mean 2, which is choice D.

Independent analysis by ActuaryProof · reviewed July 31, 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

Read the common exponential rate from the density and denote the earliest of the three losses by W.

\lambda=\frac{1}{6}

W=\min(X,Y,Z)

Model

For independent exponential clocks, the minimum is exponential with rate equal to the sum of the component rates.

\lambda_W=\lambda_X+\lambda_Y+\lambda_Z

\lambda_W=3\left(\frac{1}{6}\right)=\frac{1}{2}

Compute

Take the reciprocal of the minimum's rate to obtain its mean.

\operatorname{E}[W]=\frac{1}{\lambda_W}

\operatorname{E}[W]=\frac{1}{1/2}=2

Answer

The expected minimum loss is 2.

\boxed{2\quad\text{(D)}}

Test the method on unseen problems

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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.

AThis reports the individual exponential rate 1/6 as though it were the requested mean. Rate and mean are reciprocals.
BThis treats the component scale as 1 and divides by three to get 1/3, omitting the actual six-unit mean of each loss.
CThis correctly adds the rates to obtain 1/2 but reports that rate instead of taking its reciprocal for E[W].
EThis reports the mean 6 of one component and ignores the reduction caused by taking the minimum of three independent variables.

Original practice · fully worked

Original variant: identify a competing termination cause

A laboratory run ends at the first of two independent events. Signal arrival has an exponential waiting time with mean 4 minutes, while a safety timer has exponential rate 0.15 per minute. The log shows that the run ended within 3 minutes. Calculate the probability that the safety timer ended the run.

A
0.150
B
0.250
C
0.375
D
0.400
E
0.625

Answer and calculation

The signal and timer rates are 0.25 and 0.15. Integrating the timer-first density through minute 3 and dividing by the probability of any termination by minute 3 cancels the common time factor, leaving 0.15/0.40=0.375 and choice C.

Setup

Convert the signal mean to a rate and add the two competing rates.

\lambda_S=\frac{1}{4}=0.25

\lambda_T=0.15

\lambda_S+\lambda_T=0.40

Model

At time t, a timer termination requires both clocks to survive to t and then the timer to ring.

f_{T\text{ first}}(t)=0.15e^{-0.40t}

\Pr(\min(S,T)\le3)=1-e^{-0.40(3)}

Compute

Integrate the timer-first density over the observed window and normalize by the window probability.

\Pr(T\text{ first},\min(S,T)\le3)=\int_0^3 0.15e^{-0.40t}\,dt

\Pr(T\text{ first}\mid\min(S,T)\le3)=\frac{(0.15/0.40)(1-e^{-1.2})}{1-e^{-1.2}}=0.375

Answer

The safety timer caused the observed termination with probability 0.375.

\boxed{0.375\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: identify a competing termination cause

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables