Worked solution · Exam P

SOA Exam P Sample Question #170 Solution

This Exam P sample reference tests Sampling Without Replacement. Enumerating the six unordered category compositions gives a total weighted absolute difference of 33 across 45 equally likely pairs. The expectation is 33/45=11/15, 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 Sampling Without Replacement question

Setup

Treat each unordered two-person subset as equally likely.

N_{\mathrm{all}}=\binom{10}{2}=45

Model

Group pairs by their two categories and attach the corresponding absolute difference.

N_{RR}=3,\ N_{DD}=1,\ N_{II}=10

N_{RD}=6,\ N_{RI}=15,\ N_{DI}=10

Compute

Same-party partisan pairs contribute distance two, mixed partisan-independent pairs contribute one, and the other pairs contribute zero.

E\lvert R-D\rvert=\frac{2(3+1)+1(15+10)}{45}

E\lvert R-D\rvert=\frac{33}{45}=\frac{11}{15}

Answer

The expected absolute imbalance is eleven fifteenths.

\boxed{\frac{11}{15}\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 1/5 is E[R-D]=2(0.3-0.2), the signed difference. Taking an absolute value prevents positive and negative imbalances from canceling.
BA value of 7/15 would have weighted numerator 21 among the 45 pairs. The RI and DI pairs alone contribute 15+10=25, even before RR and DD are included.
CThe value 3/5 is E[R]=2(3/10), the expected number from the first partisan category, not the absolute difference of two counts.
EThe value 1 is E[R+D]=2(5/10), the expected total number of partisans interviewed rather than their absolute imbalance.

Original practice · fully worked

Original variant: absolute gap between two spinner scores

Two independent fair four-section spinners have section values 0, 1, 2, and 6. Let X and Y be their respective scores. Determine the expected value of the absolute score gap |X-Y|.

A
0
B
19/16
C
9/4
D
19/8
E
3.221

Answer and calculation

The six unordered unequal value pairs have gaps summing to 19. Each occurs in two orders among 16 ordered outcomes, so E|X-Y|=2(19)/16=19/8, choice D.

Setup

All sixteen ordered spinner outcomes have equal probability.

\Pr(X=x,Y=y)=\frac1{16}

Model

Diagonal outcomes contribute zero; each unordered unequal pair appears in both orders.

\sum_{x<y}(y-x)=1+2+6+1+5+4=19

Compute

Double the unordered gap total and divide by sixteen.

E\lvert X-Y\rvert=\frac{2(19)}{16}=\frac{19}{8}=2.375

Answer

The expected absolute score gap is 19/8.

\boxed{\frac{19}{8}\quad\text{(D)}}

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How to solve this Sampling Without Replacement question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: absolute gap between two spinner scores

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables