SOA Exam P Sample Question #664. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #664 — Verified Solution

Independent solution

How to solve this Exponential Distribution question

Answer in brief

An exponential observation exceeds its own mean with probability e^{-1}. The number of exceedances in ten independent observations is binomial, and its nine-or-ten tail equals 0.000825499, matching choice B.

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)}}

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-e^{-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

Variant answer in brief

Each outage exceeds four hours with probability e^{-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)}}

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ActuaryProof. “SOA Exam P Sample Question #664 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.