SOA Exam P Sample Question #718. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #718 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

Two adjacent Poisson modes at 2 and 3 force the associated rate to equal 3, and the stated mean ratio therefore makes the other rate 12. A Poisson distribution with integer rate 12 has modes 11 and 12, so the verified answer is choice D.

Setup

Use the ratio of consecutive Poisson probabilities to characterize when neighboring counts have equal maximum probability.

p_k=e^{-\lambda}\frac{\lambda^k}{k!},\qquad \frac{p_k}{p_{k-1}}=\frac{\lambda}{k}

Model

For the distribution whose two modes are 2 and 3, those two masses must be equal.

\frac{p_3}{p_2}=\frac{\lambda_{\mathrm{low}}}{3}=1

\lambda_{\mathrm{low}}=3

Compute

Apply the four-to-one mean relationship, recalling that a Poisson mean equals its rate. For an integer rate n, the equal largest masses occur at n-1 and n.

\lambda_{\mathrm{high}}=4\lambda_{\mathrm{low}}=4(3)=12

\operatorname{Modes}(\operatorname{Poisson}(12))=\{11,12\}

Answer

The two modal counts are 11 and 12, which are listed together under choice D.

\boxed{11\text{ and }12\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

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

AThe pair 8 and 9 would be the modes for rate 9, but that rate violates the required four-to-one relationship because the lower rate is 3.
BThe values 8 and 12 come from multiplying the two original modal counts 2 and 3 by four; modal counts do not scale linearly with the mean.
CThe value 10 is the midpoint of the incorrectly scaled values 8 and 12, not a Poisson mode obtained from the rate.
EThe value 12 is the new mean and is one mode, but an integer Poisson rate also makes the preceding count 11 equally likely.

Original practice · fully worked

Original variant: network packet-failure mode

A coastal monitoring network models its hourly packet-failure count N with a Poisson distribution. During calibration, engineers determine that the probability of exactly 7 failures is three halves of the probability of exactly 6 failures. Determine the mode or modes of N.

A
6 only
B
7 only
C
9 only
D
10 only
E
10 and 11

Variant answer in brief

The consecutive-mass ratio is lambda divided by 7, so the calibration relationship gives lambda equal to 10.5. A noninteger Poisson rate has the unique mode floor(10.5) = 10, making D correct.

Setup

Write the ratio of the two consecutive Poisson masses.

\frac{\Pr(N=7)}{\Pr(N=6)}=\frac{e^{-\lambda}\lambda^7/7!}{e^{-\lambda}\lambda^6/6!}=\frac{\lambda}{7}

Model

Set the theoretical ratio equal to the observed calibration ratio.

\frac{\lambda}{7}=\frac{3}{2}

\lambda=10.5

Compute

Because the rate is not an integer, the Poisson mass increases through floor(lambda) and decreases afterward.

\frac{p_{10}}{p_9}=\frac{10.5}{10}>1,\qquad \frac{p_{11}}{p_{10}}=\frac{10.5}{11}<1

\operatorname{Mode}(N)=\lfloor 10.5\rfloor=10

Answer

The unique modal packet-failure count is 10.

\boxed{10\text{ only}\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #718 — Verified Solution.” Verified 2026-07-30.

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