Worked solution · Exam P

SOA Exam P Sample Question #413 Solution

This Exam P sample reference tests Negative Binomial Distribution. This is a truncated waiting-time expectation for the second success in independent trials. The effective success probability is 0.18, and summing the four possible stopping-time contributions gives 4.73117072, which selects choice B.

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 Negative Binomial Distribution question

Setup

A trial produces the event of interest only when both required conditions occur. Multiply their probabilities and denote the complementary probability by q.

p=(0.90)(0.20)=0.18

q=1-p=0.82

Model

Let T be the number of trials observed, stopping at the second success or at the five-trial cap. For stopping times two through four, the final trial is a success and exactly one earlier trial is a success. Reaching the cap means that the first four trials contain at most one success.

\Pr(T=k)=\binom{k-1}{1}p^2q^{k-2},\qquad k=2,3,4

\Pr(T=5)=q^4+4pq^3

Compute

Evaluate the stopping probabilities and form the finite expectation.

\Pr(T=2)=0.0324,\quad \Pr(T=3)=0.053136

\Pr(T=4)=0.06535728,\quad \Pr(T=5)=0.84910672

\operatorname{E}[T]=2(0.0324)+3(0.053136)+4(0.06535728)+5(0.84910672)

\operatorname{E}[T]=4.73117072

Answer

The expected number observed rounds to 4.73.

\boxed{4.73\quad\text{(B)}}

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.

AUsing 0.20 itself as the success probability, and thereby ignoring the other required condition, gives an expectation of 4.6752 and leads to 4.68.
CUsing 0.16 as the joint probability, from multiplying 0.20 by the wrong complementary rate 0.80, gives 4.78299392 and leads to 4.78.
DAn expectation near 4.83 would require an effective success probability near 0.14. That rate is inconsistent with the required product 0.90 × 0.20.
EAn expectation near 4.88 would require an effective success probability near 0.116. This understates the correctly computed 0.18 and makes the stopping time too long.

Original practice · fully worked

Original variant: staged inspection with changing alarm rates

A laboratory checks up to four independent sensors in a fixed order. Their alarm probabilities are 0.10, 0.20, 0.30, and 0.40, respectively. Testing stops at the first alarm; if no earlier sensor alarms, the fourth sensor is still checked. Let T be the number of sensors checked. Calculate E[T].

A
0.504
B
1.914
C
2.124
D
3.124
E
4.000

Answer and calculation

Use the tail-sum formula with the probability of reaching each stage. The expectation is 1+0.90+(0.90)(0.80)+(0.90)(0.80)(0.70)=3.124, so choice D.

Setup

A stage is reached only if every earlier sensor remains quiet.

\Pr(T\ge1)=1,\qquad \Pr(T\ge2)=0.90

\Pr(T\ge3)=(0.90)(0.80)=0.72

\Pr(T\ge4)=(0.90)(0.80)(0.70)=0.504

Model

For a positive integer-valued variable capped at four, its expectation is the sum of its four tail probabilities.

\operatorname{E}[T]=\sum_{k=1}^{4}\Pr(T\ge k)

Compute

Add the probabilities of reaching the successive checks.

\operatorname{E}[T]=1+0.90+0.72+0.504=3.124

Answer

The expected number of sensors checked is 3.124.

\boxed{3.124\quad\text{(D)}}

Keep practicing

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How to solve this Negative Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: staged inspection with changing alarm rates

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables