Worked solution · Exam P

SOA Exam P Sample Question #258 Solution

This Exam P sample reference tests Set Counting. This is a three-set membership-count problem. Membership totals show that 450 drivers have exactly two factors; adding 1,200 with exactly one, 300 with all three, and 500 with none gives 2,450 and choice C.

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 Set Counting question

Setup

Group drivers by how many of the three factors they exhibit. The three exactly-one regions contain 1,200 drivers in total.

n_1=3(400)=1200,\qquad n_3=300

Model

Summing the three individual factor counts counts an exactly-two driver twice and an all-three driver three times.

1000+1000+1000=n_1+2n_2+3n_3

Compute

Solve for the number with exactly two factors, then add the disjoint membership classes and the no-factor class.

3000=1200+2n_2+3(300)

n_2=450

N=n_1+n_2+n_3+n_0=1200+450+300+500=2450

Answer

The study contains 2,450 drivers.

\boxed{2450\quad\text{(C)}}

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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.

AThe value 2,000 is 1,200+300+500. It includes the exactly-one, all-three, and no-factor classes but omits every driver with exactly two factors.
BThe value 2,300 is obtained by including only two of the three 150-driver exactly-two regions: 1,200+2(150)+300+500. One pairwise-only region is missing.
DThe value 2,750 counts the all-three group twice after the disjoint partition: 1,200+450+2(300)+500. Each driver must appear once in the study total.
EThe value 3,500 adds the three individual factor counts to the 500 with no factors. The individual counts overlap, so exactly-two and all-three drivers are counted repeatedly.

Original practice · fully worked

Original variant: exactly one flag among flagged samples

A laboratory audits 600 samples using flags A, B, and C. The individual flag counts are 280, 250, and 220. The inclusive pairwise intersection counts are 120, 90, and 80, and 40 samples carry all three flags. Among samples bearing one or more flags, determine the proportion bearing exactly one.

A
0.167
B
0.283
C
0.340
D
0.483
E
0.580

Answer and calculation

Inclusion-exclusion gives 500 samples with at least one flag. Removing repeated pair and triple memberships gives 290 samples with exactly one flag, so the conditional probability is 290/500=0.58 and choice E.

Setup

Apply three-set inclusion-exclusion to count samples carrying at least one flag.

n(A\cup B\cup C)=280+250+220-120-90-80+40

=500

Model

Recover the exactly-one count from the individual, inclusive pairwise, and triple counts.

n_1=\sum n(A_i)-2\sum n(A_i\cap A_j)+3n(A\cap B\cap C)

Compute

Evaluate the exactly-one count and normalize it within the flagged samples.

n_1=750-2(290)+3(40)=290

\Pr(\text{exactly one}\mid A\cup B\cup C)=\frac{290}{500}=0.58

Answer

Among flagged samples, 58.0 percent carry exactly one flag.

\boxed{0.580\quad\text{(E)}}

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How to solve this Set Counting question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: exactly one flag among flagged samples

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability