Worked solution · Exam P

SOA Exam P Sample Question #146 Solution

This Exam P sample reference tests Negative Binomial Distribution. A third malfunction on day five requires exactly two malfunctions in the first four days and one on day five. Dividing that probability by 1-0.4 cubed gives 0.14769, choice C.

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

Let A be the stated stopping-day event and B the conditioning event.

A=\{T_3=5\}

B=\{\text{fewer than three malfunctions in days }1,2,3\}

Model

Event A automatically satisfies B, and its probability follows from the negative-binomial trial arrangement.

A\subseteq B

\Pr(A)=\binom42(0.4)^2(0.6)^2(0.4)=0.13824

Compute

The only way B fails is a malfunction on each of the first three days.

\Pr(B)=1-(0.4)^3=0.936

\Pr(A\mid B)=\frac{0.13824}{0.936}=0.1476923

Answer

The conditional probability rounds to 0.148.

\boxed{0.148\quad\text{(C)}}

Test the method on unseen problems

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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 0.064 is (0.4)³, the probability of three malfunctions in the first three days. That is precisely the excluded complement of the conditioning event.
BThe value 0.138 is the unconditioned probability of the third malfunction occurring on day five. It omits division by 0.936.
DThe value 0.230 results from dividing the numerator by 0.60, as though the condition were a single non-malfunction day rather than the three-day event.
EThe value 0.246 uses an incorrect conditional denominator. The applicable denominator is 1-(0.4)³ because only three early malfunctions are excluded.

Original practice · fully worked

Original variant: infer a flag rate from two stopping masses

Independent diagnostic cycles raise a flag with the same unknown probability p. A monitor opens a review upon the fourth flag. The probability that the review opens on cycle six is 1.75 times the probability that it opens on cycle five. Determine p.

A
0.20
B
0.25
C
0.30
D
0.40
E
0.70

Answer and calculation

The ratio of the two negative-binomial masses is [C(5,3)/C(4,3)](1-p)=2.5(1-p). Equating it to 1.75 gives p=0.30, choice C.

Setup

Write the two stopping-time masses with the fourth flag fixed on the final cycle.

\Pr(T_4=6)=\binom53p^4(1-p)^2

\Pr(T_4=5)=\binom43p^4(1-p)

Model

Take the ratio so the common fourth-power factor cancels.

\frac{\Pr(T_4=6)}{\Pr(T_4=5)}=\frac{\binom53}{\binom43}(1-p)=2.5(1-p)

Compute

Match the observed probability ratio and solve for the flag rate.

2.5(1-p)=1.75

1-p=0.70,\qquad p=0.30

Answer

Each diagnostic cycle raises a flag with probability 0.30.

\boxed{p=0.30\quad\text{(C)}}

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Next steps for Exam P Univariate Random Variables

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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: infer a flag rate from two stopping masses

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables