Verified solution · Exam P

SOA Exam P Sample Question #193 Solution

This Exam P sample reference tests Permutations. Among 24 assignments of four labeled items, nine are derangements with no correct placement. The complement therefore has probability 1-9/24=5/8, choice D.

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.

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Independent solution

How to solve this Permutations question

Setup

Count all assignments and use the complement event of no correct placements.

\#\{\text{all assignments}\}=4!=24

\Pr(\text{at least one match})=1-\frac{D_4}{4!}

Model

Evaluate the derangement count by inclusion–exclusion.

D_4=4!\left(1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}\right)=9

Compute

Subtract the nine no-match assignments from the full permutation space.

\Pr(\text{at least one match})=1-\frac9{24}=\frac{15}{24}=\frac58

Answer

At least one correct placement occurs with probability five eighths.

\boxed{\frac58\quad\text{(D)}}

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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 27/256 is (1/4)(3/4)³, treating placements as independent and counting one specified item as the sole match. A random permutation creates dependent placements and any of four items could match.
BThe value 1/4 is the probability that one specified item is correctly placed. It omits assignments where another item matches.
CThe value 11/24 counts only eleven favorable permutations and misses four valid assignments. Direct derangement counting shows that 24-9=15 assignments have at least one match.
EThe value 3/4 is the probability that one specified item is misplaced. It is neither the target nor its complement, because other placements remain dependent.

Original practice · fully worked

Original variant: overlapping divisibility screens

Equally likely tickets bear the integers 1 through 120. A screening rule accepts a drawn ticket when its number is divisible by 4, 6, or 10. What is the acceptance probability for a single drawing?

A
1/60
B
7/20
C
11/30
D
31/60
E
19/30

Variant answer in brief

The single divisibility counts total 62, the pairwise-overlap counts total 20, and the triple overlap contains 2 integers. Inclusion–exclusion gives 62-20+2=44 accepted integers, so the probability is 11/30, choice C.

Setup

Count the multiples for each individual divisibility test.

N_4=\frac{120}{4}=30,\qquad N_6=20,\qquad N_{10}=12

Model

Use least common multiples to count pairwise and triple overlaps.

N_{4,6}=\frac{120}{12}=10,\quad N_{4,10}=\frac{120}{20}=6,\quad N_{6,10}=\frac{120}{30}=4

N_{4,6,10}=\frac{120}{60}=2

Compute

Apply inclusion–exclusion and divide the accepted count by 120.

N_{\mathrm{accepted}}=30+20+12-10-6-4+2=44

\Pr(\mathrm{accepted})=\frac{44}{120}=\frac{11}{30}

Answer

The acceptance probability is eleven thirtieths.

\boxed{\frac{11}{30}\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #193 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Permutations question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: overlapping divisibility screens

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution