Worked 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 July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

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)}}

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.

See the cover and sample pages

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

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: first signal with a delayed backup detector

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables