Worked 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 July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

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)}}

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.

See the cover and sample pages

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Negative Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: parity from a retry-count generating function

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables