Worked solution · Exam P

SOA Exam P Sample Question #538 Solution

This Exam P sample reference tests Geometric Distribution. The trial count through the first success is geometric with success probability 0.0625. Its probability mass decreases by the factor 0.9375 at each successive trial, so the mode is 1 and choice A is correct.

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.

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Independent solution

How to solve this Geometric Distribution question

Setup

Let T count the number of independent trials up to and including the first success.

p=0.0625,\qquad T\sim\operatorname{Geometric}(p)

Model

The geometric mass at trial n consists of n-1 failures followed by one success.

\Pr(T=n)=(1-p)^{n-1}p,\qquad n=1,2,\ldots

Compute

Compare each probability with the one immediately before it.

\frac{\Pr(T=n+1)}{\Pr(T=n)}=1-p=0.9375<1

Answer

The mass function is strictly decreasing, so its largest value occurs at the first trial.

\boxed{\operatorname{mode}(T)=1\quad\text{(A)}}

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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.

BThis approximates a lower-tail quantile and then adds an extra trial, an off-by-one adjustment that is unrelated to the mode.
CThis is the median trial count, since the smallest n with 1-0.9375ⁿ at least 0.5 is 11; a median is not a mode.
DThis is the expected number of failures before success, (1-p)/p=15, rather than the most likely trial count.
EThis is the mean number of trials through success, 1/p=16, rather than the mode.

Original practice · fully worked

Original variant: decoder retry percentile

A decoder makes independent attempts until its first successful read. Let F be the number of failed attempts before the success, and suppose Pr(F at least 4)=0.2401. Let T=F+1 be the total number of attempts. Calculate the 75th percentile of T, defined as the smallest integer n for which Pr(T at most n) is at least 0.75.

A
1
B
2.333
C
3.333
D
4
E
5

Answer and calculation

Four initial failures have probability (1-p)⁴=0.2401=0.7⁴, so p=0.3. The geometric CDF is 1-0.7ⁿ; it is 0.657 at n=3 and 0.7599 at n=4, making 4 the 75th percentile and choice D correct.

Setup

Use the supplied tail event to recover the failure probability on one attempt.

\Pr(F\ge 4)=(1-p)^4=0.2401=0.7^4

p=0.3

Model

For total attempts through the first success, the cumulative probability through attempt n is one minus the probability of n consecutive failures.

\Pr(T\le n)=1-(1-p)^n=1-0.7^n

Compute

Check the two integers around the target cumulative probability.

\Pr(T\le 3)=1-0.7^3=0.657

\Pr(T\le 4)=1-0.7^4=0.7599

Answer

The cumulative probability first reaches 0.75 at four attempts.

\boxed{4\quad\text{(D)}}

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How to solve this Geometric Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: decoder retry percentile

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables