Verified solution · Exam P

SOA Exam P Sample Question #261 Solution

This Exam P sample reference tests Binomial Distribution. The sales count is Binomial(5,0.20). Complementing the zero-sale and one-sale outcomes gives 1-0.8⁵-5(0.2)(0.8)⁴=0.26272, which rounds to 0.26 and choice C.

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 Binomial Distribution question

Setup

Let S be the number of successful appointments.

S\sim\operatorname{Binomial}(5,0.20)

Model

It is shortest to complement the two disjoint outcomes with fewer than two sales.

\Pr(S\ge2)=1-\Pr(S=0)-\Pr(S=1)

Compute

Evaluate the two binomial masses and subtract them from one.

\Pr(S\ge2)=1-(0.8)^5-\binom51(0.2)(0.8)^4

=1-0.32768-0.40960=0.26272

Answer

The probability rounds to 0.26.

\boxed{0.26\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.

AThe value 0.04 is (0.20)², the probability that two specified appointments both result in sales. It ignores the other appointments and every placement of two or more sales.
BThe value 0.20 is the success probability for one appointment, not the probability of at least two successes in five.
DThe value 0.40 is 2(0.20). It adds two single-appointment probabilities without counting placements or correcting overlapping multi-sale outcomes.
EThe value 0.74 is Pr(S≤1)=0.73728 rounded. It is the complement of the requested event.

Original practice · fully worked

Original variant: dispersion of a threshold quality bonus

Eight independent quality checks each pass with probability 0.25. A technician receives a bonus of 120 credits if at least two checks pass and receives no bonus otherwise. Calculate the standard deviation of the bonus.

A
0.3671
B
44.05
C
57.84
D
75.95
E
95.47

Variant answer in brief

The bonus event has probability 1-0.75⁸-8(0.25)(0.75)⁷=0.6329193. The bonus is 120 times a Bernoulli variable with that parameter, so its standard deviation is 120sqrt(q(1-q))=57.84, choice C.

Setup

Let Q be the number of passing checks and q the probability that the bonus is earned.

Q\sim\operatorname{Binomial}(8,0.25)

Model

Complement zero or one pass to obtain the bonus probability.

q=\Pr(Q\ge2)=1-(0.75)^8-\binom81(0.25)(0.75)^7

q=0.6329193115234375

Compute

The bonus B is a scaled Bernoulli variable.

\operatorname{Var}(B)=120^2q(1-q)

\operatorname{SD}(B)=120\sqrt{q(1-q)}=57.8410526822

Answer

The standard deviation rounds to 57.84 credits.

\boxed{57.84\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #261 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: dispersion of a threshold quality bonus

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution