Worked solution · Exam P

SOA Exam P Sample Question #590 Solution

This Exam P sample reference tests Geometric Distribution. This is a geometric stopping-time calculation. A round continues with probability (1/4)³+(3/4)³=7/16, so exactly five rounds requires four continuations followed by a stop: (7/16)⁴(9/16)=0.020608, making choice C correct.

Independent analysis by ActuaryProof · reviewed July 30, 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

In one round, continuation occurs only when all three outcomes match. Add the all-first-outcome and all-second-outcome probabilities.

r=\left(\frac14\right)^3+\left(\frac34\right)^3=\frac7{16}

Model

Exactly five rounds means that each of the first four rounds continues and the fifth round stops.

\Pr(T=5)=r^4(1-r)

Compute

Substitute the one-round continuation and stopping probabilities.

\Pr(T=5)=\left(\frac7{16}\right)^4\left(\frac9{16}\right)

\Pr(T=5)=\frac{21609}{1048576}=0.0206079483\ldots

Answer

The probability rounds to 0.0206.

\boxed{0.0206\quad\text{(C)}}

Test the method on unseen problems

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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 treats continuation as a prescribed rare matching pattern rather than allowing both possible all-matching outcomes.
BThis over-multiplies the rare all-first-outcome event instead of using the full one-round continuation probability 7/16.
DThis is close to an off-by-one calculation with only three continuation rounds before the stopping round.
EThis is close to r²(1-r), the probability of stopping on round three rather than round five.

Original practice · fully worked

Original variant: two-state incident response

An incident starts at level 0. During the first response cycle it is resolved with probability 0.3 or advances to level 1 with probability 0.7. Once at level 1, each later cycle resolves it with probability 0.6 or leaves it at level 1 with probability 0.4. Calculate the probability that the incident is resolved exactly on cycle 4.

A
0.0288
B
0.0672
C
0.0960
D
0.1680
E
0.2800

Answer and calculation

Resolution on cycle 4 requires advancement on cycle 1, two further unresolved level-1 cycles, and resolution on cycle 4. The probability is 0.7(0.4)²(0.6)=0.0672, so choice B is correct.

Setup

Identify the unique state path that permits resolution on the fourth cycle.

0\longrightarrow 1\longrightarrow 1\longrightarrow 1\longrightarrow \text{resolved}

Model

The first transition has probability 0.7, the next two cycles must remain unresolved, and the final cycle must resolve.

\Pr(T=4)=0.7(0.4)(0.4)(0.6)

Compute

Multiply the transition probabilities along the required path.

\Pr(T=4)=0.7(0.4)^2(0.6)=0.0672

Answer

The incident resolves on cycle 4 with probability 0.0672.

\boxed{0.0672\quad\text{(B)}}

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How to solve this Geometric Distribution question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: two-state incident response

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability