Worked solution · Exam P

SOA Exam P Sample Question #412 Solution

This Exam P sample reference tests Order Statistics. This is the CDF of a sample maximum from three independent uniform observations. All three observations must fall below the threshold, giving 0.7 cubed or 0.343 and selecting choice D.

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 Order Statistics question

Setup

For one observation on the stated interval, the probability of falling below seven is the fraction of the interval below that threshold.

\Pr(X_i<7)=\frac7{10}=0.7

Model

The sample maximum is below seven exactly when every selected observation is below seven.

\{\max(X_1,X_2,X_3)<7\}=\bigcap_{i=1}^{3}\{X_i<7\}

Compute

Use independence to multiply the three identical probabilities.

\Pr(\max(X_1,X_2,X_3)<7)=(0.7)^3=0.343

Answer

The probability that the maximum is below seven is 0.343.

\boxed{0.343\quad\text{(D)}}

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 0.027 is 0.3 cubed, the probability that all three observations exceed seven.
BThe value 0.081 is three times 0.3 cubed. Multiplying the all-above event by three double-counts a single outcome pattern.
CThe value 0.189 is 3(0.7)(0.3)², the probability that exactly one of the three observations is below seven.
EThe value 0.441 is 3(0.7)²(0.3), the probability that exactly two observations are below seven.

Original practice · fully worked

Original variant: median priority in a sample without replacement

Seven maintenance tickets carry distinct priority scores 1 through 7. Three tickets are selected uniformly without replacement, and M is the median of their three scores. Calculate Pr(M≥5).

A
1/35
B
12/35
C
13/35
D
135/343
E
22/35

Answer and calculation

The sample median is at least five exactly when at least two selected tickets come from scores 5, 6, and 7. Hypergeometric counting gives (12+1)/35=13/35, so choice C.

Setup

Partition the tickets into three high scores, 5 through 7, and four lower scores. Every three-ticket subset is equally likely.

\#\{\text{all samples}\}=\binom73=35

Model

For the middle selected score to be at least five, the sample must contain exactly two or exactly three high-score tickets.

\#\{\text{exactly two high}\}=\binom32\binom41=12

\#\{\text{exactly three high}\}=\binom33\binom40=1

Compute

Add the two disjoint favorable counts and divide by all samples.

\Pr(M\ge5)=\frac{12+1}{35}=\frac{13}{35}=0.371428\ldots

Answer

The probability that the sample median is at least five is 13/35.

\boxed{\frac{13}{35}\quad\text{(C)}}

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How to solve this Order Statistics question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: median priority in a sample without replacement

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables