Worked solution · Exam P

SOA Exam P Sample Question #303 Solution

This Exam P sample reference tests Geometric Distribution. This is a geometric waiting-time probability. One experiment succeeds on either of two all-matching outcomes, so its success probability is 1/4; two failures followed by a success have probability (3/4)²(1/4)=9/64, or 0.140625 and 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

Count the two qualifying configurations among the eight equally likely outcomes of one experiment.

p=\frac{2}{2^3}=\frac14,\qquad q=1-p=\frac34

Model

A first success on experiment three requires failures on experiments one and two, then success on experiment three.

\Pr(T=3)=q^2p

Compute

Multiply the independent experiment-level probabilities.

\Pr(T=3)=\left(\frac34\right)^2\left(\frac14\right)=\frac9{64}=0.140625

Answer

The probability rounds to 0.141.

\boxed{0.141\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.

AThe value 0.047 is 3/64=(3/4)(1/4)². It uses one failure followed by two successes instead of two failures followed by the first success.
BThe value 0.070 is 9/128, obtained by retaining only one of the two terminal all-matching success configurations while still using the combined failure probability.
DThe value 0.188 is 3/16=(3/4)(1/4), the probability that the first success occurs on experiment two.
EThe value 0.422 is 27/64=(3/4)³, the probability that all first three experiments fail.

Original practice · fully worked

Original variant: which independent uploader finishes first

Uploaders A and B make one independent transmission attempt per round until their own first success. Each A attempt succeeds with probability 0.25, and each B attempt succeeds with probability 0.50. Calculate the probability that uploader A succeeds in a strictly earlier round than uploader B.

A
0.125
B
0.200
C
0.250
D
0.333
E
0.800

Answer and calculation

If A first succeeds in round k while B remains unsuccessful through that round, the probability is 0.25(0.50)(0.75 × 0.50)⁽ᵏ⁻¹⁾. Summing the geometric series gives 0.125/(1-0.375)=0.20, so choice B.

Setup

Let T_A and T_B be the two independent geometric completion rounds.

p_A=0.25,\quad q_A=0.75,\qquad p_B=0.50,\quad q_B=0.50

Model

For A to win in round k, both uploaders must fail each earlier round, then A succeeds while B fails in round k.

\Pr(T_A=k,T_B>k)=p_Aq_B(q_Aq_B)^{k-1}

Compute

Sum over every possible winning round.

\Pr(T_A<T_B)=\sum_{k=1}^{\infty}0.125(0.375)^{k-1}

=\frac{0.125}{1-0.375}=0.20

Answer

Uploader A finishes strictly first with probability 0.200.

\boxed{0.200\quad\text{(B)}}

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: which independent uploader finishes first

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables