Verified solution · Exam P

SOA Exam P Sample Question #322 Solution

This Exam P sample reference tests Negative Binomial Distribution. This is a negative-binomial waiting-time variance calculation. Reaching the third target face is the sum of three independent geometric waiting times with success probability 1/6, so the variance is 3(5/6)/(1/6)²=90 and choice E.

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 Negative Binomial Distribution question

Setup

Treat the designated die result as a Bernoulli success and identify the required number of successes.

r=3,\qquad p=\frac16,\qquad q=1-p=\frac56

Model

The total waiting time to the third success is negative binomial. Equivalently, it is the sum of three independent geometric waiting times.

X=G_1+G_2+G_3,\qquad \operatorname{Var}(G_i)=\frac{q}{p^2}

Compute

Add the three identical geometric variances.

\operatorname{Var}(X)=\frac{rq}{p^2}

=\frac{3(5/6)}{(1/6)^2}=90

Answer

The variance of the total roll count is 90.

\boxed{90\quad\text{(E)}}

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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 5/12 equals 3(1/6)(5/6), applying the Bernoulli variance formula to three trials. The number of trials is random, and the geometric variance divides by p squared.
BThe value 18/25 results from swapping success and failure probabilities in the negative-binomial variance: 3(1/6)/(5/6)².
CThe value 15 is the expected number of failures before the third success, rq/p, not the variance of the total number of rolls.
DThe value 30 is q/p squared, the variance of the wait to one success. The total wait contains three independent success cycles.

Original practice · fully worked

Original variant: parity from a retry-count generating function

The number N of retry messages generated during a network handshake has probability-generating function G_N(s)=[0.40/(1-0.60s)]⁵. Calculate the probability that N is even.

A
0.00098
B
0.25000
C
0.49951
D
0.50049
E
0.99902

Variant answer in brief

For an integer count, G_N(-1) is the even-probability minus the odd-probability. Here G_N(-1)=(1/4)⁵=1/1024; combining that difference with probabilities summing to one gives P(N even)=1025/2048=0.50049, choice D.

Setup

Separate the generating series into its even-power and odd-power contributions.

P_E=\Pr(N\text{ is even}),\qquad P_O=\Pr(N\text{ is odd})

P_E+P_O=G_N(1)=1

Model

Substitution of negative one changes the sign of every odd-power probability and preserves every even-power probability.

G_N(-1)=P_E-P_O

Compute

Evaluate the generating function at negative one, then solve the two equations for the even probability.

G_N(-1)=\left(\frac{0.40}{1+0.60}\right)^5=\left(\frac14\right)^5=\frac1{1024}

P_E=\frac{1+G_N(-1)}2=\frac{1025}{2048}=0.50048828125

Answer

The probability of an even retry count rounds to 0.50049.

\boxed{0.50049\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #322 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Negative Binomial Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: parity from a retry-count generating function

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution