Worked solution · Exam P

SOA Exam P Sample Question #295 Solution

This Exam P sample reference tests Independence. Complement the requested event with exactly three and exactly four purchases. Those upper-count probabilities are 0.15 and 0.0125, so the probability of at most two purchases is 0.8375 and choice D.

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 Independence question

Setup

Let N count purchases and complement the only two counts above two.

\Pr(N\le2)=1-\Pr(N=3)-\Pr(N=4)

Model

Exactly three purchases arise either when the fourth customer purchases with exactly two of the first three, or when the first three purchase and the fourth does not.

\Pr(N=3)=\binom32(0.5)^2(0.5)(0.1)+(0.5)^3(0.9)

=0.0375+0.1125=0.15

Compute

Calculate all four purchases and take the complement.

\Pr(N=4)=(0.5)^3(0.1)=0.0125

\Pr(N\le2)=1-0.15-0.0125=0.8375

Answer

The probability rounds to 0.84.

\boxed{0.84\quad\text{(D)}}

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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 is P(N=2)=0.375. It includes only one count rather than all outcomes with zero, one, or two purchases.
BThis is P(N≤1)=0.1125+0.35=0.4625. It incorrectly excludes the allowed two-purchase outcomes.
CThis is P(N≥2)=1-P(N≤1)=0.5375, which reverses the direction of the count condition.
EThis is the probability that the fourth customer does not purchase. That single-customer event does not determine the total count.

Original practice · fully worked

Original variant: identity check with checksum alternatives

An archive release requires an identity check to pass and at least one of two checksum checks to pass. The three checks operate independently. Their pass probabilities are 0.64 for identity, 0.72 for the first checksum, and 0.85 for the second checksum. Calculate the probability that the archive is released.

A
0.042
B
0.392
C
0.613
D
0.640
E
0.958

Answer and calculation

The chance that at least one checksum passes is 1-(0.28)(0.15)=0.958. Multiplying by the independent identity-pass probability gives 0.61312, so choice C.

Setup

Use a complement to calculate availability of the checksum portion.

\Pr(\text{at least one checksum})=1-(1-0.72)(1-0.85)

=1-(0.28)(0.15)=0.958

Model

Release also requires the independent identity check, so intersect the two required conditions.

\Pr(\text{release})=0.64(0.958)

Compute

Evaluate the product.

0.64(0.958)=0.61312

Answer

The release probability rounds to 0.613.

\boxed{0.613\quad\text{(C)}}

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How to solve this Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: identity check with checksum alternatives

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability