Worked solution · Exam P

SOA Exam P Sample Question #212 Solution

This Exam P sample reference tests Geometric Distribution. Writing q for the probability that one patient does not have the disease gives r=q cubed. Conditioning on reaching patient four leaves eight additional failures before patient twelve, so the answer is q to the eighth power=r⁽⁸⁄³⁾, choice C.

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

Let p be the common disease probability and let q=1-p be the probability of continuing past one patient.

\Pr(T\ge n)=q^{n-1}

Model

Express the supplied probability r and the requested conditional tail in powers of q.

r=\Pr(T\ge4)=q^3

\Pr(T\ge12\mid T\ge4)=\frac{q^{11}}{q^3}=q^8

Compute

Eliminate q using r=q cubed.

q=r^{1/3}

q^8=\left(r^{1/3}\right)^8=r^{8/3}

Answer

The conditional probability is r to the eight-thirds power.

\boxed{r^{8/3}\quad\text{(C)}}

Test the method on unseen problems

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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 expression r⁽¹¹⁄³⁾=q¹¹ is the unconditional probability of reaching patient twelve. It does not divide by the event of reaching patient four.
BThe expression r cubed=q⁹ counts nine additional failures. From the fourth tested patient to the twelfth requires eight additional failures.
DThe expression r squared=q⁶ counts only six additional failures after the conditioning point.
EThe expression r⁽¹⁄³⁾=q is the probability of only one further failure, not eight.

Original practice · fully worked

Original variant: a geometric doubling score after a checkpoint

Independent attempts continue until a success. The probability that at least three attempts are required is 0.16. After learning that the first two attempts failed, let K be the number of further failures before success. A score is defined as 2ᴷ. Calculate the expected score.

A
0.600
B
1.587
C
1.667
D
3.000
E
5.000

Answer and calculation

Two initial failures have probability 0.16, so the failure probability is 0.40 and success probability is 0.60. Memorylessness gives P(K=k)=0.60(0.40)ᵏ, and summing 2ᵏ times this mass yields 3, choice D.

Setup

Let q be the per-attempt failure probability and recover it from the two known failures.

\Pr(T\ge3)=q^2=0.16

q=0.40,\qquad p=1-q=0.60

Model

Memorylessness makes the additional failure count geometric on zero, one, two, and so on.

\Pr(K=k)=pq^k=0.60(0.40)^k,\qquad k=0,1,2,\ldots

Compute

Weight each score by its geometric probability and sum the convergent series.

\operatorname{E}[2^K]=\sum_{k=0}^{\infty}2^k(0.60)(0.40)^k

\operatorname{E}[2^K]=0.60\sum_{k=0}^{\infty}(0.80)^k=\frac{0.60}{1-0.80}=3

Answer

The expected doubling score is 3.

\boxed{3.000\quad\text{(D)}}

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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: a geometric doubling score after a checkpoint

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables