Worked solution · Exam P

SOA Exam P Sample Question #201 Solution

This Exam P sample reference tests Binomial Distribution. For one subject, the stopping probability is q=3p²(1-p)+p³. Reaching subject n requires n-1 nonstopping outcomes followed by one stopping outcome, so the probability is (1-q)ⁿ⁻¹q, choice B.

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 Binomial Distribution question

Setup

A stop occurs when the within-subject positive count is two or three.

K\sim\operatorname{Binomial}(3,p)

q=\Pr(K\ge 2)

Model

Add the two disjoint stopping cases, then view successive subjects as independent Bernoulli stopping trials.

q=\binom{3}{2}p^2(1-p)+\binom{3}{3}p^3

q=3p^2(1-p)+p^3

Compute

Exactly n tests means n-1 continuations and then a stop.

\Pr(T=n)=(1-q)^{n-1}q

=\left[1-3p^2(1-p)-p^3\right]^{n-1}\left[3p^2(1-p)+p^3\right]

Answer

The expression matches choice B.

\boxed{\left[1-3p^2(1-p)-p^3\right]^{n-1}\left[3p^2(1-p)+p^3\right]\quad\text{(B)}}

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.

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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.

AThis expression uses 3p²(1-p) as the stopping probability and omits the p³ case in which all three indicators are positive.
CThis reverses the geometric roles, using one continuation and n-1 stopping outcomes. Only the final tested subject should be a stopping outcome.
DThis is the correct geometric expression multiplied by n. The stopping position is fixed at n, so there are not n possible placements.
EThis tracks first positives for the three indicators separately. The stopping rule is a within-subject count, not an ordering of indicator-specific first hits.

Original practice · fully worked

Original variant: exactly one rejected batch

A workshop examines five independent batches. Each batch contains three modules, and each module independently fails inspection with probability 0.20. A batch is rejected when at least two of its modules fail. What is the chance that the rejection count among the five batches equals one?

A
0.0670
B
0.1040
C
0.3351
D
0.4659
E
0.5775

Answer and calculation

A batch is rejected with probability q=3(0.2)²(0.8)+(0.2)³=0.104. A Binomial(5,q) count equals one with probability 5q(1-q)⁴=0.335147, choice C.

Setup

First find the rejection probability for a single batch.

q=\Pr(\text{two failures})+\Pr(\text{three failures})

Model

The number of rejected batches is binomial because the five batches are independent.

q=\binom{3}{2}(0.2)^2(0.8)+(0.2)^3=0.104

R\sim\operatorname{Binomial}(5,0.104)

Compute

Evaluate the probability mass at one rejection.

\Pr(R=1)=\binom{5}{1}(0.104)(0.896)^4

=0.3351470355

Answer

Exactly one rejected batch has probability 0.3351.

\boxed{0.3351\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: exactly one rejected batch

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables