Worked solution · Exam P

SOA Exam P Sample Question #470 Solution

This Exam P sample reference tests Geometric Distribution. The probability law is geometric on the number of failures before the first success, with success probability 1/5. Conditioning on survival past one merely shifts the variable by two, so its variance remains (4/5)/(1/5)²=20 and choice C is correct.

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 Geometric Distribution question

Setup

Recognize the probability law as a geometric failure count with success probability p=1/5 and failure probability q=4/5.

\Pr(X=x)=p q^x,\qquad x=0,1,2,\ldots

p=\frac15,\qquad q=\frac45

Model

Given that the failure count exceeds one, at least two failures have occurred. The additional failure count has the original geometric law.

(X-2\mid X>1)\overset{d}=X

\operatorname{Var}(X\mid X>1)=\operatorname{Var}(X)

Compute

Use the variance formula for a geometric failure count.

\operatorname{Var}(X)=\frac{q}{p^2}

\operatorname{Var}(X\mid X>1)=\frac{4/5}{(1/5)^2}=20

Answer

The conditional variance is 20.

\boxed{20\quad\text{(C)}}

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.

AA finite value such as 13 can result from truncating the geometric tail after only a few terms. The conditional law still has infinite support, so tail truncation understates its variance.
BThis is the square of the unconditional geometric mean, (q/p)²=4²=16, not a variance.
DThis adds the two-unit conditioning shift to the variance 20. Adding a constant changes a mean but never a variance.
EThis adds the unconditional mean 4 to the variance 20. The memoryless residual has the same variance, not variance plus mean.

Original practice · fully worked

Original variant: residual inspection cost after repeated misses

Independent inspections find a defect with probability 0.25 on each attempt. Three unsuccessful inspections have already occurred. Let R be the number of additional unsuccessful inspections before the first detection, and let the remaining labor cost be C=10+2R dollars. Calculate Var(C).

A
12 dollars²
B
24 dollars²
C
36 dollars²
D
48 dollars²
E
96 dollars²

Answer and calculation

Past misses do not change the geometric residual count. Thus Var(R)=0.75/0.25²=12, and the coefficient 2 multiplies variance by 2², giving Var(C)=48 and choice D.

Setup

Describe the remaining miss count using the geometric failure-count model.

p=0.25,\qquad q=0.75

R\sim\operatorname{GeometricFailures}(p)

Model

Independence makes the three completed misses irrelevant to the future count, while the affine cost transformation scales variance by the square of its slope.

\operatorname{Var}(R)=\frac{q}{p^2}

\operatorname{Var}(10+2R)=4\operatorname{Var}(R)

Compute

Compute the residual-count variance and then apply the cost multiplier.

\operatorname{Var}(R)=\frac{0.75}{0.25^2}=12

\operatorname{Var}(C)=4(12)=48

Answer

The remaining labor-cost variance is 48 square dollars.

\boxed{48\ \text{dollars}^2\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Geometric Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: residual inspection cost after repeated misses

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables