Verified solution · Exam P

SOA Exam P Sample Question #211 Solution

This Exam P sample reference tests Geometric Distribution. Inside the condition, the first count is zero or one. Independence and geometric tails give favorable joint probability 4/243 and conditioning probability 8/9, so the conditional probability is 1/54=0.0185, choice E.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Geometric Distribution question

Setup

Let M and N be the independent first- and second-period counts and record the geometric tail.

\Pr(N=n)=\frac{2}{3^{n+1}},\qquad \Pr(N\ge k)=\left(\frac13\right)^k

\Pr(M<2)=\frac23+\frac29=\frac89

Model

For M equal to zero, the second count must be at least four; for M equal to one, it must be at least three.

\Pr(M+N>3,\,M<2)=\Pr(M=0)\Pr(N\ge4)+\Pr(M=1)\Pr(N\ge3)

Compute

Evaluate the two independent branches and normalize by the conditioning event.

\Pr(M+N>3,\,M<2)=\frac23\frac1{81}+\frac29\frac1{27}=\frac4{243}

\Pr(M+N>3\mid M<2)=\frac{4/243}{8/9}=\frac1{54}=0.0185185

Answer

The requested conditional probability rounds to 0.0185.

\boxed{0.0185\quad\text{(E)}}

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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 0.0062 is 1/162, obtained by shifting both required second-period thresholds one count too high, as if the total had to exceed four rather than three.
BThe value 0.0123 is 1/81. It uses the same second-period threshold of four for both M=0 and M=1, overlooking that N=3 is sufficient when M=1.
CThe value 0.0139 is 1/72. Multiplying it by the conditioning probability gives 1/81, which omits one quarter of the true favorable joint mass 4/243.
DThe value 0.0165 is 4/243, the favorable joint probability before division by P(M<2)=8/9.

Original practice · fully worked

Original variant: equal positive geometric counts

Two independent counters have the same distribution, with P(N=n)=3/4⁽ⁿ⁺¹⁾ for n=0,1,2,... . Given that at least one counter records a positive value, calculate the probability that the two recorded values are equal.

A
0.0375
B
0.0625
C
0.0857
D
0.4375
E
0.6000

Variant answer in brief

Equal values inside the condition must be a common positive integer. Their joint probability is a geometric series totaling 3/80, and the conditioning probability is 7/16, giving 3/35=0.0857, choice C.

Setup

Let X and Y be the two counts and exclude their joint zero outcome.

\Pr(X+Y>0)=1-\Pr(X=0,Y=0)=1-\left(\frac34\right)^2=\frac7{16}

Model

Under the condition, equality occurs only when both counters show the same positive integer.

\Pr(X=Y,\,X+Y>0)=\sum_{n=1}^{\infty}\left(\frac{3}{4^{n+1}}\right)^2

Compute

Sum the geometric series and divide by the probability of a positive total.

\sum_{n=1}^{\infty}\frac{9}{16^{n+1}}=\frac{9/256}{1-1/16}=\frac3{80}

\Pr(X=Y\mid X+Y>0)=\frac{3/80}{7/16}=\frac3{35}=0.0857143

Answer

The conditional equality probability rounds to 0.0857.

\boxed{0.0857\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #211 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Geometric Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: equal positive geometric counts

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution