Worked solution · Exam P

SOA Exam P Sample Question #570 Solution

This Exam P sample reference tests Counting Methods. Conditioning on three damaged items makes all C(10,3)=120 subsets equally likely. The favorable category counts are 30, 6, and 3, totaling 39, so the probability is 39/120=0.325 and choice C is 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 Counting Methods question

Setup

After conditioning on the total damaged count, use equally likely subsets of three items.

N_{\text{all}}=\binom{10}{3}=120

Model

Partition favorable subsets by the category of the third damaged item.

\begin{aligned}N_{1F,1P,1U}&=\binom21\binom31\binom51=30,\\N_{1F,2P}&=\binom21\binom32=6,\qquad N_{2F,1P}=\binom22\binom31=3.\end{aligned}

Compute

Add the disjoint favorable counts and divide by the total.

N_{\text{fav}}=30+6+3=39

\Pr(\text{at least one F and one P})=\frac{39}{120}=0.325

Answer

The required conditional probability is 0.325.

\boxed{0.325\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 applies inclusion-exclusion but omits the add-back intersection: (120-C(8,3)-C(7,3))/120=29/120=0.241667.
BThis counts only subsets with one item from each of the three categories, giving 30/120=0.250 and omitting favorable subsets with no uninsured item.
DThis multiplies the separate probabilities of at least one fully insured and at least one partially insured as if those events were independent, obtaining (64/120)(85/120)=0.377778.
EThis selects one full item, one partial item, and then any of eight remaining items, giving 2x3x8/120=0.400 but double-counting subsets containing two full or two partial items.

Original practice · fully worked

Original variant: conditioned display selection

A curator has 5 science volumes, 4 history volumes, and 3 art volumes available for a four-book display. Every four-book subset is equally likely. Given that the display contains at least one art volume, calculate the probability that it contains exactly two science volumes.

A
0.303
B
0.316
C
0.325
D
0.407
E
0.424

Answer and calculation

There are C(12,4)-C(9,4)=369 displays with an art volume. Choosing two science volumes and two others with at least one art volume gives C(5,2)[C(7,2)-C(4,2)]=150, so the conditional probability is 150/369=0.4065 and choice D.

Setup

Count the displays satisfying the conditioning event.

N_{A\ge1}=\binom{12}{4}-\binom94=495-126=369

Model

Choose two science volumes, then choose two nonscience volumes with at least one art volume.

N_{2S,A\ge1}=\binom52\left(\binom72-\binom42\right)

Compute

Evaluate the favorable count and divide by the conditional sample space.

N_{2S,A\ge1}=10(21-6)=150

\Pr(2S\mid A\ge1)=\frac{150}{369}=\frac{50}{123}=0.4065040650

Answer

The conditional probability rounds to 0.407.

\boxed{0.407\quad\text{(D)}}

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How to solve this Counting Methods question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: conditioned display selection

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability