Worked solution · Exam P

SOA Exam P Sample Question #106 Solution

This Exam P sample reference tests Competing Geometric Waiting Times. At the first round with any claim, the favorable outcome has probability 0.08 and all terminating outcomes total 0.28, so the race probability is 2/7=0.2857, choice A.

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.

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Independent solution

How to solve this Competing Geometric Waiting Times question

Setup

Compare the two geometric waiting times one round at a time. If neither examiner has a claim, the same race restarts on the next round.

\Pr(\text{neither claim})=(0.9)(0.8)=0.72

Model

Rahul finishes strictly first only when Rahul has a claim and the other examiner does not in the first round containing at least one claim.

\Pr(\text{Rahul only claims})=(0.1)(0.8)=0.08

Compute

A round terminates the race with probability 0.28, and the favorable Rahul-only outcome has probability 0.08. Their ratio is two sevenths.

\Pr(T_R<T_T)=\frac{0.08}{1-0.72}=\frac27=0.285714

Answer

Rahul examines fewer policies with probability 0.2857.

\boxed{0.2857\quad\text{(A)}}

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.

CThe value one third treats the three terminating outcomes—Rahul only, the other examiner only, and both—as equally likely, although their probabilities differ.
DThe value 0.3571 uses Rahul's marginal claim probability as the numerator and therefore counts simultaneous claims as Rahul finishing first.
EThe value 0.4000 discards simultaneous-claim rounds instead of treating them as terminating ties.

Original practice · fully worked

Original variant: which inspector finds a defect first

Two inspectors test independent streams of parts. Each part tested by inspector A is defective with probability 0.15, while each tested by inspector B is defective with probability 0.25. They test simultaneously until their first defects. Find the probability A stops first.

A
0.2500
B
0.3103
C
0.3625
D
0.4138
E
0.6375

Answer and calculation

A-only success has probability 0.15(0.75)=0.1125, while a round terminates with probability 1-0.85(0.75)=0.3625. Their ratio is 0.3103, choice B.

Setup

In a simultaneous test round, neither inspector finds a defect with probability 0.6375. After such a round, the same race restarts.

q=(0.85)(0.75)=0.6375

Model

Inspector A stops first only when A finds a defect and B does not in the first round containing at least one defect.

p_{\mathrm{A\ only}}=(0.15)(0.75)=0.1125

Compute

The first terminating round has probability 0.3625, while the favorable A-only outcome has probability 0.1125. Their ratio is 0.310345.

\Pr(T_A<T_B)=\frac{0.1125}{1-0.6375}=0.310345

Answer

Inspector A stops first with probability 0.3103.

\boxed{0.3103\quad\text{(B)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Competing Geometric Waiting Times question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: which inspector finds a defect first

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables