Worked solution · Exam P

SOA Exam P Sample Question #566 Solution

This Exam P sample reference tests Poisson Distribution. For a Poisson count with mean 1.2, the unconditional upper tail from 3 is 0.1205129 and the positive-count probability is 0.6988058. Their ratio is 0.1724555, so choice D is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 Poisson Distribution question

Setup

Express the requested probability as a conditional Poisson tail.

\Pr(X\ge3\mid X\ge1)=\frac{\Pr(X\ge3)}{\Pr(X\ge1)}

Model

Use complements to evaluate the numerator and denominator.

\Pr(X\ge3)=1-e^{-1.2}\left(1+1.2+\frac{1.2^2}{2}\right)

\Pr(X\ge1)=1-e^{-1.2}

Compute

Divide the unconditional upper tail by the probability of a positive count.

\Pr(X\ge3)=0.1205129012

\Pr(X\ge1)=0.6988057881

\Pr(X\ge3\mid X\ge1)=0.1724554995

Answer

The conditional probability rounds to 0.17.

\boxed{0.17\quad\text{(D)}}

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.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis uses the single mass P(X=3)=0.0867439 without conditioning; its nearest listed value is 0.10.
BThis reports the unconditional tail P(X≥3)=0.1205129 and omits division by P(X≥1).
CThis divides the tail by the unrelated complement P(X≤2)=0.8794871, obtaining 0.1370263, whose nearest listed value is 0.15.
EThis reports the unconditional mass P(X=2)=0.2168598, an off-by-one threshold with no conditioning.

Original practice · fully worked

Original variant: mean on active days

A backup service records a Poisson number R of automatic restarts per day. Logs show that 40% of days have no restarts. Among days with at least one restart, calculate the expected number of restarts.

A
0.400
B
0.600
C
0.916
D
1.527
E
2.291

Answer and calculation

The zero probability gives λ=-ln(0.4)=0.9162907. Since R is zero off the conditioning event, E[R|R≥1]=λ/0.6=1.5271512, so choice D is correct.

Setup

Recover the Poisson mean from the observed zero-count probability.

e^{-\lambda}=0.40

\lambda=-\ln(0.40)=0.9162907319

Model

Use the fact that R contributes zero outside the positive-count event.

E[R]=E[R\mid R\ge1]\Pr(R\ge1)

\Pr(R\ge1)=0.60

Compute

Divide the unconditional Poisson mean by the positive-day probability.

E[R\mid R\ge1]=\frac{0.9162907319}{0.60}

E[R\mid R\ge1]=1.5271512198

Answer

An active day averages about 1.527 restarts.

\boxed{1.527\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: mean on active days

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables