Worked solution · Exam P

SOA Exam P Sample Question #505 Solution

This Exam P sample reference tests Poisson Distribution. Seven independent daily Poisson counts combine into a Poisson count with mean 7(0.20)=1.40. Its mass at three is 0.112777, which rounds to 0.113 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 Poisson Distribution question

Setup

Let X_i denote the count on day i and let N be the total across the seven-day horizon.

X_i\sim\operatorname{Poisson}(0.20)

N=\sum_{i=1}^{7}X_i

Model

Independent Poisson variables add to another Poisson variable whose mean is the sum of the component means.

N\sim\operatorname{Poisson}(7\cdot0.20)=\operatorname{Poisson}(1.40)

\Pr(N=k)=e^{-1.40}\frac{1.40^k}{k!}

Compute

Evaluate the aggregate Poisson mass at the requested count.

\Pr(N=3)=e^{-1.40}\frac{1.40^3}{3!}

\Pr(N=3)=0.1127770115\ldots

Answer

The requested probability rounds to 0.113.

\boxed{0.113\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.

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 can result from applying a one-step Poisson recurrence only once and retaining the daily rate: exp(−1.4)(0.2/2)=0.02466, rather than using the full weekly term 1.4³⁄³!.
BThis aggregates only six days, producing exp(−1.2)(1.2)³⁄³!=0.086744, which rounds to 0.087.
DThis replaces exp(−1.4) by the invalid shortcut 1/(1+1.4); then 1.4³/[6(2.4)]=0.19056, which rounds to 0.191.
EThis sets the Poisson mean equal to the target count instead of aggregating the daily means: exp(−3)3³⁄³!=0.224042.

Original practice · fully worked

Original variant: zero-truncated photon count

A photon counter records a Poisson number N in each calibration window. The audit log gives P(N=2)/P(N=1)=0.75 but does not list the mean. For a window known to contain at least one photon, calculate the probability that it contains exactly two.

A
0.223
B
0.251
C
0.323
D
0.431
E
0.750

Answer and calculation

The adjacent-mass ratio identifies the Poisson mean as 1.5. Conditioning the two-count mass on a nonzero window gives 0.323119, so choice C is correct.

Setup

Use the ratio of adjacent Poisson masses to recover the unreported mean.

\frac{\Pr(N=2)}{\Pr(N=1)}=\frac{\lambda}{2}=0.75

\lambda=1.50

Model

Condition the two-count mass on the event that the window is nonempty.

\Pr(N=2\mid N\ge1)=\frac{\Pr(N=2)}{1-\Pr(N=0)}

\Pr(N=2)=e^{-1.5}\frac{1.5^2}{2}

Compute

Evaluate the numerator and the zero-truncation denominator.

\Pr(N=2\mid N\ge1)=\frac{0.2510214302}{1-0.2231301601}

\Pr(N=2\mid N\ge1)=0.3231190314\ldots

Answer

The conditional probability rounds to 0.323.

\boxed{0.323\quad\text{(C)}}

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: zero-truncated photon count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables