Verified solution · Exam P

SOA Exam P Sample Question #145 Solution

This Exam P sample reference tests Geometric Distribution. A positive admission count is geometric with parameter 2/3. Enumerating the four ways that five employees can generate at most two admissions gives probability 0.78279, choice E.

Independent analysis by ActuaryProof · reviewed August 4, 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

Translate the cost threshold into a count threshold and recover the geometric parameter.

20000N<50000\iff N\le2

E[K]=\frac1p=1.5\iff p=\frac23

Model

For one employee, record the probabilities needed for aggregate counts zero, one, and two.

q_0=0.8

q_1=0.2\left(\frac23\right)=\frac2{15}

q_2=0.2\left(\frac13\right)\left(\frac23\right)=\frac2{45}

Compute

Add the disjoint aggregate configurations with total admission count at most two.

\Pr(N\le2)=q_0^5+5q_1q_0^4+5q_2q_0^4+\binom52q_1^2q_0^3

\Pr(N\le2)=0.32768+0.2730667+0.0910222+0.0910222

\Pr(N\le2)=0.7827911

Answer

The required probability rounds to 0.78.

\boxed{0.78\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.

AA value near 0.41 results from retaining the zero-admission case and one two-admission configuration while omitting the much larger one-admission contribution.
BThe value 0.46 is already below P(N=0)+P(N=1)=0.60075, so it necessarily leaves out eligible outcomes with zero or one total admission.
CThe value 0.58 results from using geometric parameter 1/3. A geometric count on 1,2,... with mean 1.5 has parameter 2/3.
DThe value 0.69 omits one of the two distinct 0.09102 contributions: one employee with two admissions or two employees with one each.

Original practice · fully worked

Original variant: low tail of a compound alert count

The number N of incidents reaching a network hub in one hour is Poisson with mean 0.8. Each incident independently produces a positive geometric number of alerts with mean 2, and alert counts from different incidents are independent. Find the probability that the hub receives at most one alert during the hour.

A
0.17973
B
0.44933
C
0.55067
D
0.62906
E
0.80879

Variant answer in brief

At most one alert means either no incidents, or one incident producing exactly one alert. These probabilities sum to exp(-0.8)(1+0.8/2)=0.62906, choice D.

Setup

Recover the geometric success parameter from its mean.

E[K]=\frac1p=2\iff p=\frac12

\Pr(K=1)=\frac12

Model

Because every incident creates at least one alert, list the only cases yielding total zero or one.

\{S=0\}=\{N=0\}

\{S=1\}=\{N=1,K_1=1\}

Compute

Evaluate the two disjoint compound-count cases.

\Pr(S\le1)=e^{-0.8}+e^{-0.8}(0.8)\left(\frac12\right)

\Pr(S\le1)=1.4e^{-0.8}=0.6290605

Answer

The low-tail probability is approximately 0.62906.

\boxed{0.62906\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #145 — 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: low tail of a compound alert count

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution