SOA Exam P Sample Question #625. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #625 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

Independent Poisson increments make the observed early count irrelevant to the remaining interval except for fixing the number still needed. The final three years must contain exactly two events, whose probability is 0.22404, so choice E is correct.

Setup

The conditioning information fixes eight events in the completed seven-year portion. Reaching ten by the end therefore requires two more events.

10-8=2

Model

Counts from disjoint years are independent. The count in the remaining three-year interval is Poisson with mean equal to three times the annual mean.

R\sim\operatorname{Poisson}(3)

Compute

Evaluate the Poisson probability of exactly two events in that remaining interval.

\Pr(R=2)=e^{-3}\frac{3^2}{2!}

\Pr(R=2)=0.2240418077\ldots

Answer

The conditional probability rounds to 0.224.

\boxed{0.224\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

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

AUsing e^(-3) calculates no events in the remaining three years. Two additional events, not zero, are required to reach the specified total.
BThis is approximately the unconditional probability of ten events over all ten years, using a Poisson mean of ten. It discards the observed seven-year count instead of conditioning on it.
CThis is approximately the probability of eight events in the first seven years, the event already supplied as information. The task is to evaluate what happens afterward.
DThis is close to a two-event probability under a one-year Poisson mean. The remaining exposure lasts three years, so its mean is three rather than one.

Original practice · fully worked

Original variant: downlink timing within a pass

Telemetry downlinks arrive according to a homogeneous Poisson process. During a four-hour satellite pass, exactly six downlinks are recorded. Conditional on that total, calculate the probability that exactly two of the downlinks occurred during the first hour.

A
0.234
B
0.251
C
0.297
D
0.356
E
0.466

Variant answer in brief

Given six arrivals in four hours, each arrival independently falls in the first hour with probability 1/4. The resulting binomial probability is 0.29663, so choice C is correct.

Setup

Condition on the observed total of six arrivals over the complete four-hour pass.

N(4)=6

Model

Conditional on the total, homogeneous Poisson arrival times are uniformly distributed across the pass. Each arrival therefore lands in the first one-quarter of the interval with probability one-quarter.

K\mid N(4)=6\sim\operatorname{Binomial}\!\left(6,\frac14\right)

Compute

Evaluate the conditional binomial mass at two.

\Pr(K=2\mid N(4)=6)=\binom{6}{2}\left(\frac14\right)^2\left(\frac34\right)^4

\Pr(K=2\mid N(4)=6)=0.2966308594\ldots

Answer

The conditional probability rounds to 0.297.

\boxed{0.297\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #625 — Verified Solution.” Verified 2026-07-30.

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