SOA Exam P Sample Question #632. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #632

The number of covered occurrences is binomial with five trials and probability 0.5. Capping the count at three gives an expected payment of 30(73/32) = 68.4375, so the nearest listed value is D.

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Independent solution

How to solve this Binomial Distribution question

Answer in brief

The number of covered occurrences is binomial with five trials and probability 0.5. Capping the count at three gives an expected payment of 30(73/32) = 68.4375, so the nearest listed value is D.

Setup

Let N count the covered occurrences across the five independent periods.

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

B=30\min(N,3)

Model

Group all counts of three or more at the contractual payment limit.

E[\min(N,3)]=\sum_{k=0}^{2}k\Pr(N=k)+3\Pr(N\ge3)

Compute

Use the symmetric binomial probabilities with denominator 32.

E[\min(N,3)]=\frac{5+20+3(10+5+1)}{32}=\frac{73}{32}

E[B]=30\left(\frac{73}{32}\right)=68.4375

Answer

The expected payment rounds to the listed value 68.

\boxed{68\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

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

AThis treats the three-payment limit as three ordinary expected trials, giving 3(30)(0.5), instead of applying the limit to the five-period count.
BThis undercounts the upper tail as though only two payments were available and then rounds the result.
CThis is obtained by omitting the five-occurrence case from the capped upper tail, producing 65.625 before rounding.
EThis is the uncapped expectation 30(5)(0.5), so it ignores that counts above three receive no additional payment.

Original practice · fully worked

Original variant: fabrication safety triggers

A fabrication line completes seven independent production runs. Its safety relay is triggered on any run with probability 0.25. A reliability rebate ignores the first two triggers, then pays 50 for each additional trigger, subject to at most three paid triggers. Calculate the expected rebate.

A
0.000
B
12.180
C
16.351
D
16.422
E
83.258

Variant answer in brief

For a binomial trigger count N, the number of paid triggers is min((N-2)+,3). Its expectation is the sum of the three binomial tails beginning at 3, 4, and 5, which equals 0.3270263672. Multiplication by 50 gives 16.35131836, so choice C is correct.

Setup

Let N count relay triggers over the seven independent runs and translate the two-trigger threshold and three-payment ceiling into a payment variable.

N\sim\operatorname{Binomial}(7,0.25)

R=50\min\!\left((N-2)_+,3\right)

Model

Apply the tail-sum identity to the paid-trigger count, whose possible values are zero through three.

E\!\left[\min\!\left((N-2)_+,3\right)\right]

=\Pr(N\ge3)+\Pr(N\ge4)+\Pr(N\ge5)

Compute

Evaluate the three required binomial tails and multiply their sum by the rebate per paid trigger.

\Pr(N\ge3)+\Pr(N\ge4)+\Pr(N\ge5)

=0.2435913086+0.0705566406+0.0128784180

=0.3270263672

E[R]=50(0.3270263672)=16.35131836

Answer

The expected reliability rebate rounds to 16.351.

\boxed{16.351\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #632 — Verified Solution.” Verified 2026-07-30.

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