Verified solution · Exam P

SOA Exam P Sample Question #561 Solution

This Exam P sample reference tests Combinatorial Probability. The desired selection contains one of the two distinguished objects and three of the six ordinary objects. The count ratio is C(2,1)C(6,3)/C(8,4)=4/7=0.5714, so choice E is correct.

Independent analysis by ActuaryProof · reviewed August 4, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Combinatorial Probability question

Setup

Among the eight objects, two have the distinguished property and six do not. The condition requires exactly one distinguished object in a sample of four.

K=2,\qquad N-K=6,\qquad n=4,\qquad X=1

Model

Count favorable unordered selections and divide by all unordered selections of four objects.

\Pr(X=1)=\frac{\binom{2}{1}\binom{6}{3}}{\binom{8}{4}}

Compute

Evaluate the combinations.

\binom{2}{1}\binom{6}{3}=2(20)=40

\binom{8}{4}=70

\Pr(X=1)=\frac{40}{70}=0.5714285714

Answer

The probability rounds to 0.57.

\boxed{0.57\quad\text{(E)}}

Continue without hunting through PDFs

All 718 Exam P sample solutions in syllabus order

The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.

See inside the Exam P study manual PDF

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis uses only the two choices for the distinguished object, 2/C(8,4)=2/70=0.0286, and omits the three ordinary selections.
BThis treats the selections as independent with replacement and counts only one position for the distinguished object, giving (2/8)(6/8)³=0.1055.
CThis counts only one of the two distinguished objects, C(6,3)/C(8,4)=20/70=0.2857.
DThis treats the four selections as independent draws with replacement, giving C(4,1)(2/8)(6/8)³=0.421875.

Original practice · fully worked

Original variant: flagged cards after screening

An archive box contains 10 index cards, of which 4 are flagged and 6 are standard. Three cards are drawn uniformly without replacement. Given that the draw contains at least one flagged card, calculate the probability that it contains exactly one flagged card.

A
0.1667
B
0.5000
C
0.5510
D
0.6000
E
0.8333

Variant answer in brief

There are 60 three-card selections with exactly one flag and 100 selections with at least one flag. The conditional probability is 60/100=0.60, so choice D is correct.

Setup

Count the exactly-one-flag selections and the selections admitted by the conditioning event.

N_{1}=\binom{4}{1}\binom{6}{2}=60

Model

Remove the no-flag selections from all three-card selections to obtain the conditional sample space.

N_{\geq 1}=\binom{10}{3}-\binom{6}{3}=120-20=100

Compute

Divide the favorable count by the conditioned count.

\Pr(X=1\mid X\geq 1)=\frac{60}{100}=0.60

Answer

The conditional probability is 0.6000.

\boxed{0.6000\quad\text{(D)}}

Keep practicing

Turn this solution into a complete study sequence

The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.

See the Exam P study manual PDF — $49 More Exam P General Probability practice problems Free Exam P formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #561 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.

How to solve this Combinatorial Probability question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: flagged cards after screening

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution