Verified solution · Exam P

SOA Exam P Sample Question #558 Solution

This Exam P sample reference tests Negative Binomial Distribution. The first nine trials must contain exactly two successes, and the tenth trial must succeed. This has probability C(9,2)(0.45)³(0.55)⁷=0.049943, so choice B is correct.

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 Negative Binomial Distribution question

Setup

For the third success to occur at trial ten, exactly two of the preceding nine trials must succeed.

N_9=2,\qquad \text{trial 10 is a success}

Model

Choose the locations of the first two successes and multiply by the independent trial probabilities.

\Pr(T_3=10)=\binom{9}{2}p^2(1-p)^7p

Compute

Insert the stated success probability.

\Pr(T_3=10)=\binom{9}{2}(0.45)^3(0.55)^7

\Pr(T_3=10)=0.04994348786

Answer

Rounded to the precision of the choices, the probability is 0.0499.

\boxed{0.0499\quad\text{(B)}}

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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.

AThis calculates one particular ordering, (0.45)³(0.55)⁷=0.00139, and omits the C(9,2)=36 possible locations of the first two successes.
CThis finds P(N_9=2)=C(9,2)(0.45)²(0.55)⁷=0.11099 but omits the required success on trial ten.
DThis uses C(9,3)(0.45)³(0.55)⁷=0.11653, describing three successes in the first nine followed by a failure.
EThis computes C(10,3)(0.45)³(0.55)⁷=0.16648, allowing the three successes to occupy any positions among all ten trials.

Original practice · fully worked

Original variant: expected packets until rejection limit

A network monitor classifies each arriving packet independently. A packet is rejected with probability 0.30, and monitoring stops as soon as the third rejected packet is observed. Calculate the expected total number of packets observed.

A
3.000
B
4.286
C
7.000
D
10.000
E
13.000

Variant answer in brief

The stopping time is negative binomial with three target events and target probability 0.30. Its mean is 3/0.30=10 packets, so choice D is correct.

Setup

Treat a rejection as the target event and let N be the trial on which the third rejection occurs.

r=3,\qquad q=0.30

Model

The expected trial count for the rth target event in independent Bernoulli trials is r divided by the target probability.

E[N]=\frac{r}{q}

Compute

Substitute the rejection count and rejection probability.

E[N]=\frac{3}{0.30}=10

Answer

The monitor observes ten packets on average.

\boxed{10.000\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #558 — Verified Solution.” Verified 2026-07-30.

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

How to solve this Negative Binomial Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: expected packets until rejection limit

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution