Worked solution · Exam P

SOA Exam P Sample Question #148 Solution

This Exam P sample reference tests Exponential Distribution. Summing the exponential mass over the final-quarter interval of every year produces a geometric series. Its value is 0.20525, which selects choice D.

Independent analysis by ActuaryProof · reviewed October 6, 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

Convert the mean to an exponential rate and express the target as disjoint repeated intervals.

\lambda=\frac12

A=\bigcup_{k=0}^{\infty}(k+0.75,k+1]

Model

Write the probability of each interval by subtracting two survival values.

\Pr(k+0.75<T\le k+1)=e^{-(k+0.75)/2}-e^{-(k+1)/2}

Compute

Factor the common within-year term and sum the resulting geometric series.

\Pr(A)=\sum_{k=0}^{\infty}e^{-k/2}\left(e^{-0.375}-e^{-0.5}\right)

\Pr(A)=\frac{e^{-0.375}-e^{-0.5}}{1-e^{-0.5}}=0.2052476

Answer

The probability is approximately 0.205.

\boxed{0.205\quad\text{(D)}}

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 0.081 is the probability of failure during only the final quarter of the first year, exp(−0.375)-exp(−0.5). It omits later years.
BThe value 0.088 is only slightly above the first-year contribution 0.08095. The second-year contribution alone is another 0.04910, so this value cannot include all later years.
CThe value 0.102 is about half the correct total and can result from applying an extra factor of the rate after interval probabilities have already been integrated.
EThe value 0.250 treats the failure's location within a calendar year as uniform. Exponential decay gives greater weight to earlier portions.

Original practice · fully worked

Original variant: expected completed inspection blocks

An industrial battery has an exponentially distributed operating time with mean five hours. It earns one inspection credit at the end of every uninterrupted two-hour block. Let N be the number of credits earned before the battery fails. Determine E[N].

A
0.32968
B
0.67032
C
2.03324
D
2.50000
E
3.03324

Answer and calculation

The probability of completing at least n blocks is exp(-0.4n). Summing these tail probabilities gives exp(-0.4)/(1-exp(-0.4))=2.03324, choice C.

Setup

Convert the mean to a rate and express a completed-block event as a survival event.

\lambda=\frac15=0.2

\{N\ge n\}=\{T\ge2n\}

Model

Evaluate the tail probability for every positive integer block count.

\Pr(N\ge n)=\Pr(T\ge2n)=e^{-0.4n}

Compute

Apply the tail-sum formula for a nonnegative integer-valued random variable.

E[N]=\sum_{n=1}^{\infty}\Pr(N\ge n)

E[N]=\sum_{n=1}^{\infty}e^{-0.4n}=\frac{e^{-0.4}}{1-e^{-0.4}}=2.0332448

Answer

The battery earns approximately 2.03324 completed-block credits on average.

\boxed{2.03324\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: expected completed inspection blocks

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables