Verified solution · Exam P

SOA Exam P Sample Question #249 Solution

This Exam P sample reference tests Exponential Distribution. The minimum exceeds a threshold only when both independent losses exceed it. Its survival function is therefore exp(-2t), identifying an exponential variable with mean 1/2 and choice B.

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

Let M denote the smaller of the two independent losses.

M=\min(X_1,X_2)

\Pr(X_i>t)=e^{-t},\qquad t\ge0

Model

Translate survival of the minimum into simultaneous survival of both losses.

\Pr(M>t)=\Pr(X_1>t,X_2>t)

=e^{-t}e^{-t}=e^{-2t}

Compute

Recognize the survival function as exponential with rate two.

M\sim\operatorname{Exponential}(\text{rate }2)

E[M]=\frac12

Answer

The expected smaller loss is 0.50.

\boxed{0.50\quad\text{(B)}}

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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.

AThe value 0.25 is Var(M)=1/(2 squared). It reports the variance of the exponential minimum instead of its expected value.
CThe value 0.75 is one half of E[max(X1,X2)]=1.50. Dividing the expected larger loss by the number of observations does not produce the expected smaller loss.
DThe value 1.00 is the mean of either loss viewed alone. It ignores the second independent opportunity to produce an earlier loss.
EThe value 1.50 is E[max(X1,X2)]=1+1/2 for two unit-mean exponentials. It selects the larger order statistic rather than the smaller one.

Original practice · fully worked

Original variant: first signal with a delayed backup detector

A primary fault detector begins operating at time zero and has an exponential signal time with mean 4 hours. If no signal has occurred by hour 1, an independent backup detector is activated; its signal time after activation is exponential with mean 2 hours. The response occurs at the first detector signal. Calculate the expected response time.

A
1.333
B
1.923
C
2.333
D
3.000
E
4.000

Variant answer in brief

Before hour 1 only the primary detector can signal; afterward the two residual rates add. Integrating the piecewise survival function gives 1.923197912 hours, so choice B.

Setup

Write the response-time survival function before and after backup activation.

\Pr(T>t)=e^{-t/4},\qquad 0\le t<1

\Pr(T>t)=e^{-t/4}e^{-(t-1)/2},\qquad t\ge1

Model

Use the tail-integral formula for a nonnegative waiting time.

E[T]=\int_0^1e^{-t/4}\,dt+\int_1^\infty e^{-t/4}e^{-(t-1)/2}\,dt

Compute

Evaluate the pre-activation contribution and the competing-rates tail.

E[T]=4(1-e^{-1/4})+\frac{e^{-1/4}}{1/4+1/2}

=1.9231979118\ldots

Answer

The expected response time is approximately 1.923 hours.

\boxed{1.923\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #249 — 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: first signal with a delayed backup detector

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution