Worked solution · Exam P

SOA Exam P Sample Question #656 Solution

This Exam P sample reference tests Binomial and Multinomial Distributions. Each of the two independent outcomes avoids the target category with probability 0.9. Therefore, the probability that the target occurs at least once is 1 − 0.9² = 0.19, so the answer is D.

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 Binomial and Multinomial Distributions question

Setup

For one outcome, the target category has probability 0.1, so its complement has probability 0.9.

P(\text{not target})=1-0.1=0.9

Model

Use independence to calculate the probability that neither of the two outcomes falls in the target category.

P(\text{no target in two outcomes})=(0.9)^2

Compute

Take the complement of the event that the target never occurs.

P(\text{at least one target})=1-(0.9)^2=0.19

Answer

The required probability is 0.19.

\boxed{0.19\quad\text{(D)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

30 original multivariate 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 halves the single-outcome target probability even though two independent opportunities increase the chance of seeing the target.
BThis adds only part of the second opportunity and omits valid outcomes.
CThis starts from the exact-one probability 0.18 and incorrectly subtracts the both-target probability 0.01.
EThis applies the same complement calculation to a category with probability 0.2 rather than the requested category with probability 0.1.

Original practice · fully worked

Original variant: component inspection outcomes

Each of three independently inspected components is classified as pass with probability 0.80, manual review with probability 0.15, or quarantine with probability 0.05. Calculate the probability that at least one component is sent for manual review.

A
0.150
B
0.325
C
0.386
D
0.450
E
0.614

Answer and calculation

A component avoids manual review with probability 0.85. All three avoid review with probability 0.85³, so the requested probability is 1 − 0.85³ = 0.385875, which rounds to choice C.

Setup

Combine the two classifications that are not manual review.

P(\text{not review})=0.80+0.05=0.85

Model

Independence makes the probability that all three components avoid review the product of their individual probabilities.

P(\text{no reviews})=(0.85)^3

Compute

Take the complement to include one, two, or three manual-review outcomes.

P(\text{at least one review})=1-(0.85)^3=0.385875

Answer

The probability rounds to 0.386.

\boxed{0.386\quad\text{(C)}}

Keep practicing

Next steps for Exam P Multivariate Random Variables

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How to solve this Binomial and Multinomial Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: component inspection outcomes

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables