Worked solution · Exam P

SOA Exam P Sample Question #664 Solution

This Exam P sample reference tests Exponential Distribution. An exponential observation exceeds its own mean with probability exp(−1). The number of exceedances in ten independent observations is binomial, and its nine-or-ten tail equals 0.000825499, matching choice B.

Independent analysis by ActuaryProof · reviewed July 30, 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

For an exponential variable, divide the threshold by the mean in the survival function. Here that ratio is one.

\Pr(X>20000)=\exp\!\left(-\frac{20000}{20000}\right)=e^{-1}

Model

Let N count observations above the threshold in the independent sample.

N\sim\operatorname{Bin}\!\left(10,e^{-1}\right)

\Pr(N\ge9)=\Pr(N=9)+\Pr(N=10)

Compute

Evaluate the two possible binomial outcomes and combine them.

\Pr(N\ge9)=\binom{10}{9}e^{-9}(1-e^{-1})+e^{-10}

\Pr(N\ge9)=\frac{10e-9}{e^{10}}=0.0008254987\ldots

Answer

The required binomial tail is represented by choice B.

\boxed{\Pr(N\ge9)=\frac{10e-9}{e^{10}}\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.

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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.

AThis reverses the coefficients that arise after factoring the two binomial tail terms. The factor must be 10-9/e, not 9-10/e.
CThis uses the nonexceedance probability 1-exp(−1) as though it were the exceedance probability and also applies an incompatible extra scaling.
DThis is the probability of exactly nine exceedances. It omits the additional qualifying outcome in which all ten observations exceed the threshold.
EThis raises the nonexceedance probability to the tenth power and attaches a factor of nine; it is neither an exact binomial mass nor the requested tail.

Original practice · fully worked

Original variant: long-duration sensor outages

Outage durations for a remote sensor are independent and exponentially distributed with mean 4 hours. During six outages selected at random, calculate the probability that at least five last longer than 4 hours.

A
0.002479
B
0.004259
C
0.006738
D
0.025555
E
0.028034

Answer and calculation

Each outage exceeds four hours with probability exp(−1). Adding the binomial masses for five and six exceedances gives 0.0280339, so choice E is correct.

Setup

The threshold equals the exponential mean, so the individual survival probability has a simple form.

p=\Pr(T>4)=e^{-4/4}=e^{-1}

Model

Let K count long outages among the six independent observations.

K\sim\operatorname{Bin}(6,e^{-1})

\Pr(K\ge5)=\Pr(K=5)+\Pr(K=6)

Compute

Add the exactly-five and all-six probabilities.

\Pr(K\ge5)=6e^{-5}(1-e^{-1})+e^{-6}

\Pr(K\ge5)=0.0280339211\ldots

Answer

The probability that at least five outages are long is approximately 0.028034.

\boxed{0.028034\quad\text{(E)}}

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: long-duration sensor outages

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables