Worked solution · Exam P

SOA Exam P Sample Question #238 Solution

This Exam P sample reference tests Geometric Distribution. Writing q=1-p, the joint calibration gives (1-q squared)²=0.0441, so q squared=0.79 and p=0.1111806. Then F(1,5)=p(1-q⁵)=0.0495073, 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 q be the probability of avoiding injury on one attempt. One participant finishes within n attempts with probability one minus q to the nth power.

q=1-p

\Pr(T\le n)=1-q^n

Model

Independence factors the joint CDF calibration at two attempts.

F(2,2)=\left(1-q^2\right)^2=0.0441

1-q^2=0.21

Compute

Recover the per-attempt injury probability and evaluate the requested rectangle.

q=\sqrt{0.79}=0.8888194,\qquad p=1-q=0.1111806

F(1,5)=\left(1-q\right)\left(1-q^5\right)

F(1,5)=0.0495073

Answer

The joint probability rounds to 0.0495.

\boxed{0.0495\quad\text{(C)}}

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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.

AUsing one attempt for both participants gives F(1,1)=p squared=0.01236, whose nearest listed value is 0.0093. It replaces the five-attempt limit by one.
BReusing the two-attempt calibration horizon for the second participant gives F(1,2)=p(1-q squared)=0.02335, nearest 0.0216.
DUsing six attempts instead of five for the second participant gives F(1,6)=p(1-q to the sixth)=0.05636, nearest 0.0551.
EThe value 0.1112 is the recovered one-attempt injury probability p. It omits the second participant's five-attempt event.

Original practice · fully worked

Original variant: strict winner in a geometric race

Two repair bots work independently in repeated rounds and stop after their first successful round. Bot A has success chance 0.30 in a round, while Bot B has success chance 0.20. Determine how likely it is that Bot A's first successful round precedes Bot B's.

A
0.1364
B
0.3000
C
0.3182
D
0.5455
E
0.6818

Answer and calculation

For A to win in round k, both bots must fail the first k-1 rounds, then A must succeed while B fails. Summing 0.24(0.56)⁽ᵏ⁻¹⁾ gives 0.24/(1-0.56)=6/11=0.5455, choice D.

Setup

Identify the probabilities of no success by either bot and of an A-only success in one round.

\Pr(\text{neither succeeds})=0.70(0.80)=0.56

\Pr(\text{A only succeeds})=0.30(0.80)=0.24

Model

An A win in round k requires k-1 rounds with no success by either bot followed by an A-only success.

\Pr(T_A=k,T_B>k)=0.56^{k-1}(0.24)

Compute

Sum the geometric series over all possible winning rounds.

\Pr(T_A<T_B)=\sum_{k=1}^{\infty}0.24(0.56)^{k-1}

=\frac{0.24}{1-0.56}=\frac6{11}=0.5454545

Answer

Bot A succeeds strictly before Bot B with probability about 0.5455.

\boxed{0.5455\quad\text{(D)}}

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How to solve this Geometric Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: strict winner in a geometric race

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables