Verified solution · Exam P

SOA Exam P Sample Question #451 Solution

This Exam P sample reference tests Exponential Distribution. This problem asks for the first order statistic of three shifted exponential lifetimes. The minimum is 5 plus an exponential variable of rate 3, so its mean is 5+1/3=5.33 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

Write each component lifetime as a fixed threshold plus an independent exponential waiting time.

Y_i=5+E_i,\qquad E_i\sim\operatorname{Exp}(1)

Model

The system lifetime is the earliest of the three component lifetimes.

T=\min(Y_1,Y_2,Y_3)=5+\min(E_1,E_2,E_3)

\min(E_1,E_2,E_3)\sim\operatorname{Exp}(1+1+1)

Compute

An exponential variable with rate 3 has mean 1/3.

E[T]=5+\frac{1}{3}=5.333333\ldots

Answer

The expected system lifetime is 5.33.

\boxed{5.33\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.

AAdding 1/5 to the shift treats the shift value as an exponential rate. The waiting-time rate is 1 for each component.
CThe minimum of three rate-one exponentials has mean 1/3, not 2/3.
DThis is 5+1, the mean lifetime of one component; a series system fails earlier than a typical component.
EThis adds more than one mean exponential waiting time and moves in the wrong direction for a minimum.

Original practice · fully worked

Original variant: source and timing of the first alert

Three independent monitoring channels A, B, and C produce alerts after exponential waiting times with rates 0.2, 0.3, and 0.5 per hour, respectively. Calculate the probability that channel C produces the first alert and that this first alert occurs within one hour.

A
0.1264
B
0.1896
C
0.3161
D
0.5000
E
0.6321

Variant answer in brief

The first-alert time has total rate 1.0, and channel C accounts for one-half of that rate. Thus the requested joint probability is 0.5(1-exp(-1))=0.3161 and choice C.

Setup

Let T be the first alert time. The three independent exponential rates add.

\Lambda=0.2+0.3+0.5=1.0

T\sim\operatorname{Exp}(1.0)

Model

For channel C to be first at time t, C must alert at t while the other two channels survive past t.

f_{C\text{ first}}(t)=0.5e^{-0.5t}e^{-0.2t}e^{-0.3t}

f_{C\text{ first}}(t)=0.5e^{-t}

Compute

Integrate the joint density over the first hour.

\Pr(C\text{ first},T\le1)=\int_0^1 0.5e^{-t}\,dt

=0.5(1-e^{-1})=0.3160602794\ldots

Answer

The requested joint probability is about 0.3161.

\boxed{0.3161\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #451 — 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: source and timing of the first alert

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution