Verified solution · Exam P

SOA Exam P Sample Question #203 Solution

This Exam P sample reference tests Discrete Order Statistics. Fixing x as the second order statistic requires one selected value below x and two above it. Dividing (x-1)C(12-x,2) by C(12,4) simplifies to (x-1)(11-x)(12-x)/990, choice B.

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.

Read the solution Open official source

Independent solution

How to solve this Discrete Order Statistics question

Setup

When the second order statistic equals x, the sample contains x itself, one smaller value, and two larger values.

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

Model

All four-element subsets of the twelve labels are equally likely.

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

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

Compute

Expand the combinations and simplify.

p(x)=\frac{(x-1)(12-x)(11-x)/2}{495}

p(x)=\frac{(x-1)(11-x)(12-x)}{990}

Answer

This formula is valid on the stated support and matches choice B.

\boxed{p(x)=\frac{(x-1)(11-x)(12-x)}{990}\quad\text{(B)}}

Continue without hunting through PDFs

All 718 Exam P sample solutions in syllabus order

The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.

See inside the Exam P study manual PDF

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis has the correct three factors but denominator 495. It counts the two larger selections as ordered and is twice the correct probability.
CThe factors (x-1)(x-2)(12-x)/990 select two values below x and one above, which describes x as the third order statistic.
DThis combines the third-order-statistic factors with denominator 495, additionally omitting the factor 1/2 for choosing the two lower values.
EThe three factors above x resemble choosing x as the minimum, but denominator 495 treats the three larger selections as ordered and omits their factor 3!.

Original practice · fully worked

Original variant: continuous second-order threshold

Four independent response times are uniformly distributed between 0 and 1 second. Let Y be the second smallest response time. Calculate the probability that Y is at most 0.40 seconds.

A
0.0256
B
0.3456
C
0.4752
D
0.5248
E
0.8704

Variant answer in brief

The second response is at most 0.40 exactly when at least two of the four times are at most 0.40. A Binomial(4,0.4) tail gives 0.5248, choice D.

Setup

Count how many of the four response times fall at or below the threshold.

K=\#\{i:X_i\le0.40\}

Model

Uniformity and independence make K binomial, and the second order statistic crosses the threshold when K is at least two.

K\sim\operatorname{Binomial}(4,0.40)

\{Y\le0.40\}=\{K\ge2\}

Compute

Use the complement containing zero or one qualifying response.

\Pr(K\ge2)=1-(0.6)^4-4(0.4)(0.6)^3

=1-0.1296-0.3456=0.5248

Answer

The threshold probability is 0.5248.

\boxed{0.5248\quad\text{(D)}}

Keep practicing

Turn this solution into a complete study sequence

The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.

See the Exam P study manual PDF — $49 More Exam P Univariate Random Variables practice problems Free Exam P formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #203 — 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: continuous second-order threshold

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution