Worked solution · Exam P

SOA Exam P Sample Question #15 Solution

This Exam P sample reference tests Convolution. There are eight feasible weekly splits of seven claims, and every convolution term equals 2⁻⁹. Their sum is 1/64, 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.

Read the solution Open official source

Independent solution

How to solve this Convolution question

Setup

Condition on the first week's count n. For a total of seven, the second week's count must be 7-n, with n ranging from 0 through 7.

\Pr(N_1+N_2=7)=\sum_{n=0}^{7}\Pr(N_1=n)\Pr(N_2=7-n)

Model

Independence gives a discrete convolution. Under the supplied mass function, every one of the eight feasible split probabilities simplifies to 2 to the power -9.

\Pr(N_1=n)\Pr(N_2=7-n)=2^{-(n+1)}2^{-(8-n)}=2^{-9}

Compute

Summing the eight equal convolution terms gives 8 × 2 to the power -9, or 2 to the power -6.

\Pr(N_1+N_2=7)=8\cdot2^{-9}=2^{-6}

Answer

Therefore the probability of seven claims in total is 1/64, corresponding to choice D.

\boxed{\frac1{64}\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.

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 1/256 equals two of the eight convolution terms, so it counts only two feasible weekly splits.
BThe value 1/128 equals four of the eight equal terms and therefore counts only half of the feasible splits.
CThe value 7/512 counts seven split values. The endpoints n=0 and n=7 are both feasible, giving eight terms.
EThe value 1/32 equals sixteen terms of size 1/512, double the actual number of feasible weekly splits.

Original practice · fully worked

Original variant: convolution of two uniform counters

Two independent counters each display one of the integers 0 through 5 with equal probability. Determine the probability that their displayed values sum to 7.

A
1/18
B
1/12
C
5/36
D
1/6
E
1/9

Answer and calculation

The feasible ordered pairs are (2,5), (3,4), (4,3), and (5,2). Four of 36 equally likely pairs give 1/9, choice E.

Setup

The ordered pair of counter values is uniformly distributed over 36 pairs in the six-by-six sample space.

(X,Y)\in\{0,1,2,3,4,5\}^2

Model

Enumerate the ordered pairs whose coordinates sum to seven while both coordinates remain between zero and five.

X+Y=7\Longrightarrow (X,Y)\in\{(2,5),(3,4),(4,3),(5,2)\}

Compute

There are four favorable ordered pairs, each with probability 1/36, so the total probability is 4/36=1/9.

\Pr(X+Y=7)=4\left(\frac16\right)^2=\frac19

Answer

The probability that the displayed values sum to seven is 1/9, which is choice E.

\boxed{\frac19\quad\text{(E)}}

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 Convolution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: convolution of two uniform counters

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables