Verified solution · Exam P

SOA Exam P Sample Question #179 Solution

This Exam P sample reference tests Discrete Order Statistics. For x=3,...,11, fixing x as the second largest selected value requires two choices below x and one above it. Dividing C(x-1,2)(12-x) by C(12,4) simplifies to (x-1)(x-2)(12-x)/990, choice A.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Discrete Order Statistics question

Setup

When the second largest value equals x, the sample contains x, two lower values, and one higher value. Such a selection is possible for x from 3 through 11.

N_x=\binom{x-1}{2}\binom{12-x}{1}

x\in\{3,4,\ldots,11\}

Model

All four-element subsets of the twelve labels have equal probability.

N_{\mathrm{all}}=\binom{12}{4}=495

p(x)=\frac{N_x}{N_{\mathrm{all}}}

Compute

Expand the combination and simplify the denominator.

p(x)=\frac{[(x-1)(x-2)/2](12-x)}{495}

p(x)=\frac{(x-1)(x-2)(12-x)}{990}

Answer

On its positive support, the resulting probability function matches choice A.

\boxed{p(x)=\frac{(x-1)(x-2)(12-x)}{990},\quad x=3,\ldots,11\quad\text{(A)}}

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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.

BThis has the correct factors but denominator 495. It treats the choice of two lower values as ordered and doubles the probability.
CThese factors correspond to one lower value and two higher values, the pattern for a second-smallest statistic, and the denominator also misses a factor 2.
DThe expression with one lower and two higher values over 990 is the probability function for the second smallest, not the second largest.
EThe three upper-side factors treat x as the minimum and use an incorrect combination factor; choosing three larger values requires division by 3!.

Original practice · fully worked

Original variant: expected exponential response order

Four independent devices have exponential response times with mean 0.5 seconds. Let T be the time at which the second device responds. Calculate E[T].

A
1/8
B
1/6
C
7/24
D
1/2
E
25/24

Variant answer in brief

With four rate-2 clocks, the first arrival spacing has mean 1/8 and the next spacing has mean 1/6. Their sum is 7/24, choice C.

Setup

Each response clock has rate 2.

\lambda=\frac{1}{0.5}=2

Model

Exponential order-statistic spacings are independent; their rates equal the number of clocks still running times λ.

E[T_{(1)}]=\frac{1}{4\lambda}

E[T_{(2)}-T_{(1)}]=\frac{1}{3\lambda}

Compute

Add the first two mean spacings.

E[T_{(2)}]=\frac18+\frac16=\frac{7}{24}=0.2916667

Answer

The expected second response time is 7/24 seconds.

\boxed{\frac{7}{24}\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #179 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Discrete Order Statistics question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: expected exponential response order

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution