Worked solution · Exam P

SOA Exam P Sample Question #659 Solution

This Exam P sample reference tests Negative Binomial Distribution. Let X be the trial number of the second success. The negative-binomial probabilities for X=10, 11, and 12 are 0.046695, 0.036318, and 0.027965; their sum is 0.110978, which rounds to choice D.

Independent analysis by ActuaryProof · reviewed July 30, 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

Model the outcomes as independent Bernoulli trials with success probability 0.30, and let X mark the trial on which success number two occurs.

p=0.30,\qquad 1-p=0.70

Model

For the second success to occur on trial x, exactly one of the first x-1 trials succeeds and trial x also succeeds.

\Pr(X=x)=\binom{x-1}{1}(0.30)^2(0.70)^{x-2},\qquad x=2,3,\ldots

Compute

Evaluate the three mutually exclusive arrival times and add their probabilities.

\Pr(X=10)=9(0.30)^2(0.70)^8=0.0466948881

\Pr(X=11)=10(0.30)^2(0.70)^9=0.0363182463

\Pr(X=12)=11(0.30)^2(0.70)^{10}=0.0279650497

\Pr(10\le X\le12)=0.1109781841

Answer

The requested three-trial window has probability approximately 0.1110.

\boxed{0.1110\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 0.0280 is only P(X=12), so it retains the final endpoint but omits the valid arrival times 10 and 11.
BThe value 0.0363 is only P(X=11), the middle term of the required three-term sum.
CThe value 0.0856 results from the off-by-one sum of x(0.30)²(0.70)⁽ˣ⁻¹⁾ for x=10 through 12, advancing both the eligible-position count and the failure count.
EThe value 0.1223 uses x rather than x-1 as the number of possible locations for the first success, even though the final trial is already fixed as success number two.

Original practice · fully worked

Original variant: corrupted telemetry frames

An automated observatory independently labels each incoming telemetry frame as corrupted with probability 0.20. Its emergency channel opens when the third corrupted frame is observed. Calculate the probability that the channel opens on frame 8, 9, or 10.

A
0.058720
B
0.139335
C
0.174168
D
0.224395
E
0.524288

Answer and calculation

For the third corrupted frame to arrive at position x, two of the first x-1 frames must be corrupted and frame x must also be corrupted. Summing x=8 through 10 gives 0.174168, so choice C is correct.

Setup

Let T be the position of the third corrupted frame in an independent sequence with corruption probability 0.20.

p=0.20,\qquad q=0.80

Model

An arrival at position x requires two corruptions among the preceding x-1 frames, followed by a corruption at x.

\Pr(T=x)=\binom{x-1}{2}p^3q^{x-3}

Compute

Add the three disjoint arrival-time probabilities.

\Pr(T=8)=\binom{7}{2}(0.20)^3(0.80)^5=0.05505024

\Pr(T=9)=\binom{8}{2}(0.20)^3(0.80)^6=0.058720256

\Pr(T=10)=\binom{9}{2}(0.20)^3(0.80)^7=0.0603979776

\Pr(8\le T\le10)=0.1741684736

Answer

The emergency channel opens in the stated frame window with probability about 0.174168.

\boxed{0.174168\quad\text{(C)}}

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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: corrupted telemetry frames

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables