Worked solution · Exam P

SOA Exam P Sample Question #268 Solution

This Exam P sample reference tests Binomial Distribution. Let N be the binomial count of covered years. The benefit is 20 times min(N,3); weighting its four possible payment levels gives 45.625, which rounds to choice D.

Independent analysis by ActuaryProof · reviewed July 31, 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

Model the number of years producing a covered event.

N\sim\operatorname{Binomial}(5,0.5)

B=20\min(N,3)

Model

At probability one-half, the six binomial masses have numerator row 1, 5, 10, 10, 5, 1 over 32.

\Pr(N=k)=\binom5k\frac1{32},\qquad k=0,\ldots,5

Compute

Apply the payment ceiling to the count and form the expectation.

E[B]=20\left[\frac5{32}+2\left(\frac{10}{32}\right)+3\left(\frac{16}{32}\right)\right]

=20\left(\frac{73}{32}\right)=45.625

Answer

The expected benefit rounds to 46.

\boxed{46\quad\text{(D)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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 30 is 60P(N≥3)=60(16/32). It counts the capped payment only in years-count states at or above three and discards the positive benefits when N is one or two.
BSumming payments only for N=1,2,3 and incorrectly assigning zero at N=4,5 gives [20(5)+40(10)+60(10)]/32=34.375, which rounds to 34.
CThe value 40 is 20(4)(0.5). It counts only four annual exposure periods and ignores the five-year count as well as the payment ceiling.
EThe value 50 is the uncapped expectation 20E[N]=20(5)(0.5). It pays for the fourth and fifth covered years even though no further benefits are available.

Original practice · fully worked

Original variant: expected overtime under an escalation layer

Eight independent processing tasks each require escalation with probability 0.25. The first escalation in a batch is handled during regular time. Each of the next three escalations adds 30 minutes of overtime; any further escalations are deferred. Calculate the expected overtime for the batch.

A
18.988
B
32.046
C
33.003
D
55.628
E
60.000

Answer and calculation

For the binomial escalation count N, paid overtime units are min((N-1) positive part,3). Its tail-sum expectation is P(N at least 2)+P(N at least 3)+P(N at least 4)=1.0681915, giving 32.045746 minutes and choice B.

Setup

Model the number of escalated tasks and translate the contract into paid overtime units.

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

U=\min\{(N-1)_+,3\},\qquad O=30U

Model

Use the tail-sum identity for the nonnegative integer-valued number of paid overtime units.

E[U]=\Pr(N\ge2)+\Pr(N\ge3)+\Pr(N\ge4)

Compute

Evaluate the three binomial tails and convert overtime units to minutes.

E[U]=0.6329193115+0.3214569092+0.1138153076

E[O]=30(1.0681915283)=32.04574585\ldots

Answer

The expected overtime is approximately 32.046 minutes.

\boxed{32.046\quad\text{(B)}}

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How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: expected overtime under an escalation layer

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables