SOA Exam P Sample Question #639. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #639 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

The two independent Poisson counts must fall at or below their respective coverage limits. Their cumulative probabilities are 2.5e to the negative one and 1.3e to the negative 0.3, whose product is 0.885728; this is choice B.

Setup

Let X and Y be the two annual event counts. The required outcome occurs when each count stays within its own limit.

X\sim\operatorname{Poisson}(1),\qquad Y\sim\operatorname{Poisson}(0.3)

\{\text{both limits met}\}=\{X\le2\}\cap\{Y\le1\}

Model

Independence allows the joint event probability to factor into two Poisson cumulative probabilities.

\Pr(X\le2,Y\le1)=\Pr(X\le2)\Pr(Y\le1)

Compute

Sum each Poisson mass from zero through the applicable upper bound.

\Pr(X\le2)=e^{-1}\left(1+1+\frac{1}{2}\right)=2.5e^{-1}

\Pr(Y\le1)=e^{-0.3}(1+0.3)=1.3e^{-0.3}

\Pr(X\le2,Y\le1)=(2.5e^{-1})(1.3e^{-0.3})=0.8857283274\ldots

Answer

The product matches the expression in choice B.

\boxed{(2.5e^{-1})(1.3e^{-0.3})\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

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

AThis multiplies the probabilities of landing exactly at both limits. Counts below either limit are also fully accommodated and must be included.
CThis is the complement of the probability that both limits are met, so it describes at least one count exceeding its allowance.
DThis multiplies the two exceedance probabilities and therefore requires both counts to be over their limits.
EThis is the complement of both counts exceeding their limits; it allows one count to exceed its limit and is therefore too broad.

Original practice · fully worked

Original variant: greenhouse service thresholds

During a greenhouse season, the number of pump resets is Poisson with mean 0.6, and the number of sensor recalibrations is Poisson with mean 1.4. The counts are independent. A basic service plan covers at most one reset and at most two recalibrations. Calculate the probability that every service event is covered.

A
0.020
B
0.080
C
0.268
D
0.732
E
0.980

Variant answer in brief

The two relevant Poisson cumulative probabilities are 1.6e to the negative 0.6 and 3.38e to the negative 1.4. Their independent product is 0.731893, which rounds to choice D.

Setup

Let R count pump resets and let C count sensor recalibrations during the season.

R\sim\operatorname{Poisson}(0.6),\qquad C\sim\operatorname{Poisson}(1.4)

\{\text{all events covered}\}=\{R\le1,C\le2\}

Model

Because the two counts are independent, multiply their within-threshold probabilities.

\Pr(R\le1,C\le2)=\Pr(R\le1)\Pr(C\le2)

Compute

Evaluate each short Poisson sum and then multiply.

\Pr(R\le1)=e^{-0.6}(1+0.6)=1.6e^{-0.6}

\Pr(C\le2)=e^{-1.4}\left(1+1.4+\frac{1.4^2}{2}\right)=3.38e^{-1.4}

\Pr(R\le1,C\le2)=5.408e^{-2}=0.7318932117\ldots

Answer

The service plan covers every event with probability about 0.732.

\boxed{0.732\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #639 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.