Verified solution · Exam P

SOA Exam P Sample Question #488 Solution

This Exam P sample reference tests Negative Binomial Distribution. For the third completion to occur on trial eight, the first seven trials must contain exactly two completions and trial eight must complete. The resulting probability is C(7,2)(0.25)³(0.75)⁵=0.077866…, which rounds to 0.08 and selects choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Negative Binomial Distribution question

Setup

Separate the final trial from the seven earlier independent trials. The final trial must be a success, while the earlier block must contain exactly two successes.

p=0.25,\qquad q=1-p=0.75

Model

Choose the two successful positions among the first seven, multiply by their Bernoulli probabilities, and then require success on the eighth trial.

\Pr(T_3=8)=\binom72p^2q^5p=\binom72p^3q^5

Compute

Substitute the success and failure probabilities.

\Pr(T_3=8)=21(0.25)^3(0.75)^5

\Pr(T_3=8)=0.0778656006\ldots

Answer

The probability rounds to 0.08.

\boxed{0.08\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.

AUsing an erroneous failure exponent of eight gives C(7,2)(0.25)³(0.75)⁸=0.03285, which rounds to 0.03. Only five failures occur in a valid eight-trial history with three successes.
CThis is approximately C(7,3)(0.25)³(0.75)⁵=0.1298. It places all three successes somewhere in the first seven and makes the eighth trial a failure, so the third success occurs too early.
DThis is C(8,3)(0.25)³(0.75)⁵=0.2076, the probability of exactly three successes among eight trials. It does not require the eighth trial itself to be the third success.
EThis is C(8,2)(0.25)²(0.75)⁶=0.3115, the probability of exactly two successes in all eight trials, so the third success has not yet occurred.

Original practice · fully worked

Original variant: whether a game reaches round seven

A player wins each independent round with probability 0.40. A game ends immediately when the player records a fourth win. Calculate the probability that the game lasts at least seven rounds.

A
0.17920
B
0.27648
C
0.54432
D
0.82080
E
0.95334

Variant answer in brief

The game reaches round seven exactly when the first six rounds contain at most three wins. A Binomial(6,0.40) lower-tail sum gives 0.82080, so choice D is correct.

Setup

Let W be the number of wins in the first six rounds. The game is still active for a seventh round precisely when the fourth win has not yet occurred.

W\sim\operatorname{Binomial}(6,0.40)

\{\text{game lasts at least 7 rounds}\}=\{W\le3\}

Model

Evaluate the binomial lower tail through three wins.

\Pr(W\le3)=\sum_{k=0}^{3}\binom6k(0.40)^k(0.60)^{6-k}

Compute

The four disjoint masses correspond to zero, one, two, and three wins.

\Pr(W\le3)=0.046656+0.186624+0.311040+0.276480

\Pr(W\le3)=0.820800

Answer

The probability that round seven is reached is 0.82080.

\boxed{0.82080\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #488 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

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: whether a game reaches round seven

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution