Worked solution · Exam P

SOA Exam P Sample Question #688 Solution

This Exam P sample reference tests Counting Methods. A qualifying four-item subset must have category counts 2, 1, and 1. Counting the three possible doubled categories gives 105 favorable subsets out of 210, so the probability is 1/2 and choice E.

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

All four-item subsets of the ten available items are equally likely.

N_{\mathrm{all}}=\binom{10}{4}=210

Model

To include all three categories in a sample of four, exactly one category must contribute two items and each other category must contribute one.

N_{\mathrm{good}}=\binom{5}{2}\binom{3}{1}\binom{2}{1}+\binom{5}{1}\binom{3}{2}\binom{2}{1}+\binom{5}{1}\binom{3}{1}\binom{2}{2}

Compute

Evaluate the three disjoint category-count cases and divide by the total number of subsets.

N_{\mathrm{good}}=60+30+15=105

\Pr(\text{all three categories})=\frac{105}{210}=\frac12

Answer

Exactly half of the equally likely four-item subsets contain every category.

\boxed{\frac12\quad\text{(E)}}

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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 3/100 is already smaller than the single valid 2-1-1 case with the first category doubled, whose probability is 60/210=2/7, so it severely undercounts favorable subsets.
BThe value 1/24 is also smaller than one complete valid category-count case and cannot include the three disjoint ways to choose the doubled category.
CA probability of 1/4 would correspond to 52.5 of the 210 equally likely subsets, so it cannot arise from a valid subset count and omits substantial favorable mass.
DThe value 9/25=0.36 is obtained by treating the four selections as independent draws with replacement from category probabilities 0.5, 0.3, and 0.2; the actual sampling is without replacement.

Original practice · fully worked

Original variant: seed-vault viability panel

A seed vault has 5 alpine accessions, 4 coastal accessions, and 3 desert accessions ready for a viability panel. A botanist chooses 5 distinct accessions uniformly at random. Calculate the probability that the panel contains exactly two desert accessions and at least one accession from each of the other two habitats.

A
0.1136
B
0.1515
C
0.1953
D
0.2652
E
0.3182

Answer and calculation

After choosing two of the three desert accessions, the remaining three positions split between alpine and coastal as 1+2 or 2+1. These cases give 210 of 792 panels, or 0.2652 and choice D.

Setup

Count the equally likely panels of five distinct accessions.

N_{\mathrm{all}}=\binom{12}{5}=792

Model

Choose two desert accessions. The remaining three selections must consist of one alpine and two coastal accessions or two alpine and one coastal accession.

N_{\mathrm{good}}=\binom32\left[\binom51\binom42+\binom52\binom41\right]

Compute

Evaluate the two valid alpine-coastal splits and divide by the total panel count.

N_{\mathrm{good}}=3(30+40)=210

\Pr(\text{required composition})=\frac{210}{792}=\frac{35}{132}=0.2651515\ldots

Answer

The required composition occurs with probability approximately 0.2652.

\boxed{0.2652\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: seed-vault viability panel

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability