Worked solution · Exam P

SOA Exam P Sample Question #445 Solution

This Exam P sample reference tests Set Probability. The three single-factor-only regions contain 1,260 people. Symmetry forces each pair-only region to contain 230, so the three pair-only regions contain 690; adding 320 in all three and 480 in none gives 2,750 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 Probability question

Setup

Let u_AB, u_AC, and u_BC denote the counts in exactly the indicated two-factor regions. The three exactly-one regions have equal stated size.

N_{\mathrm{exactly\ one}}=3(420)=1260

Model

Within each factor set, subtract its single-only region and the all-three region. The remainder lies in its two adjacent pair-only regions.

u_{AB}+u_{AC}=1200-420-320=460

u_{AB}+u_{BC}=460

u_{AC}+u_{BC}=460

Compute

The three equations give 230 in each pair-only region. Add the four disjoint membership categories.

u_{AB}=u_{AC}=u_{BC}=230

N_{\mathrm{exactly\ two}}=3(230)=690

N=1260+690+320+480=2750

Answer

The study contains 2,750 people.

\boxed{2750\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.

AThis adds only the exactly-one and no-factor groups, 1260+480=1740, omitting everyone with two or three factors.
BThis adds only one of the three 230-person pair-only regions: 1260+230+320+480=2290.
DThe aggregate two-factor membership count is 1380, but every exactly-two person contributes twice. Using 1380 as a person count gives 1260+1380+320+480=3440.
EThis adds the three set sizes and the no-factor group, 3(1200)+480=4080, without removing people counted in multiple sets.

Original practice · fully worked

Original variant: servers with exactly one alert class

Among 1,000 servers, alert A appears on 420, alert B on 380, and alert C on 350. The pairwise intersections contain 160, 140, and 120 servers for A and B, A and C, and B and C, respectively. Exactly 800 servers have at least one alert. Calculate the number of servers with exactly one alert class.

A
70
B
200
C
420
D
520
E
730

Answer and calculation

Inclusion-exclusion first gives 70 servers in all three alert classes. The three exactly-one regions are then 190, 170, and 160, totaling 520 and selecting choice D.

Setup

Use the known union to solve for the triple intersection.

800=(420+380+350)-(160+140+120)+N_{ABC}

Model

Recover the all-three count, then remove both pairwise overlaps from each marginal and restore the triple intersection once.

N_{ABC}=70

N_{A\text{-only}}=420-160-140+70=190

N_{B\text{-only}}=380-160-120+70=170

N_{C\text{-only}}=350-140-120+70=160

Compute

Add the three disjoint exactly-one regions.

N_{\mathrm{exactly\ one}}=190+170+160=520

Answer

There are 520 servers with exactly one alert class.

\boxed{520\quad\text{(D)}}

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How to solve this Set Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: servers with exactly one alert class

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability