Worked solution · Exam P

SOA Exam P Sample Question #286 Solution

This Exam P sample reference tests Exponential Distribution. This is a comparison of two independent exponential amounts. A one-sided ratio event has probability one-third, and the two symmetric directions are disjoint, so their total probability is two-thirds and choice E.

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

Let X and Y denote the two amounts. Their common mean is one, so each has rate one.

f_X(x)=e^{-x},\qquad \Pr(Y>t)=e^{-t},\qquad x,t\ge 0

Model

First evaluate the direction in which Y exceeds twice X. The opposite direction has the same probability, and the two strict inequalities cannot occur together.

\Pr(Y>2X)=\int_0^\infty \Pr(Y>2x)f_X(x)\,dx

Compute

Insert the exponential survival function and density, then use symmetry.

\Pr(Y>2X)=\int_0^\infty e^{-2x}e^{-x}\,dx=\int_0^\infty e^{-3x}\,dx=\frac13

\Pr(Y>2X\text{ or }X>2Y)=2\left(\frac13\right)=\frac23

Answer

The required probability is two-thirds.

\boxed{\frac23\quad\text{(E)}}

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.

AThe value 1/6 results from taking the correct one-direction probability 1/3 and multiplying by 1/2 for symmetry. Symmetry creates a second disjoint direction, so it calls for addition rather than another restriction.
BThe value 1/4 is the incorrect integral of exp(-4x), obtained by using exp(-2x) for both the conditional tail and the conditioning density. The actual conditioning density is exp(-x), so the combined exponent is three.
CThe value 1/3 is the probability of only one direction, such as Y>2X. The equally likely disjoint event X>2Y must also be included.
DThe value 1/2 is the probability that a specified amount is merely the larger one. It ignores the stronger requirement that the larger amount exceed twice the other.

Original practice · fully worked

Original variant: latency inside a directional band

A data job has independent setup latency S with exponential rate 2 and transfer latency T with exponential rate 3. A monitoring flag appears when the transfer latency is greater than the setup latency but less than three times the setup latency. Calculate the probability of a flag.

A
0.1818
B
0.2182
C
0.4000
D
0.5818
E
0.6000

Answer and calculation

The flag is the part of the tail T>S that remains after removing T>3S. Exponential tail averaging gives 2/5-2/11=12/55=0.2182, so choice B.

Setup

Write the two independent exponential models with their unequal rates.

f_S(s)=2e^{-2s},\qquad \Pr(T>t)=e^{-3t},\qquad s,t\ge0

Model

The upper-tail event T>3S is nested inside T>S, so subtract it from the broader tail.

\Pr(S<T<3S)=\Pr(T>S)-\Pr(T>3S)

Compute

Average each transfer tail over the setup density.

\Pr(T>cS)=\int_0^\infty e^{-3cs}2e^{-2s}\,ds=\frac{2}{2+3c}

\Pr(S<T<3S)=\frac25-\frac2{11}=\frac{12}{55}=0.2181818\ldots

Answer

The monitoring flag appears with probability about 0.2182.

\boxed{0.2182\quad\text{(B)}}

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: latency inside a directional band

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables