Worked solution · Exam P

SOA Exam P Sample Question #305 Solution

This Exam P sample reference tests Poisson Distribution. The two qualification conditions concern independent variables. The Poisson lower-tail probability is 4e to the minus 3, and the standardized normal threshold is 0.5; multiplying 4 exp(−3) by Φ(0.5) gives 0.1377, so choice B.

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 be the error count and T the completion time. Standardize the time threshold.

X\sim\operatorname{Poisson}(3)

Z=\frac{50-45}{10}=0.5

Model

Compute the two qualifying probabilities separately, then use the stated independence.

\Pr(X\le1)=e^{-3}+3e^{-3}=4e^{-3}

\Pr(T\le50)=\Phi(0.5)

Compute

Multiply the Poisson and normal probabilities.

\Pr(X\le1,T\le50)=4e^{-3}\Phi(0.5)

=0.1991482735(0.6914624613)=0.1377035553

Answer

The joint qualification probability rounds to 0.14.

\boxed{0.14\quad\text{(B)}}

Test the method on unseen problems

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30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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Error diagnosis

Why the other letters tempt

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

AThe value 0.10 is approximately (4 exp(−3))(0.5). It uses a standard-normal cutoff of zero instead of the correctly standardized cutoff 0.5.
CThe value 0.19 is close to 4 exp(−3), the probability of meeting the error-count condition alone. It omits the time condition.
DThe value 0.84 is close to Φ(1). It doubles the correct standardized time gap and also omits the error-count condition.
EThe value 0.89 is approximately 4 exp(−3)+Φ(0.5). The two conditions must both occur, so independence calls for a product rather than a sum.

Original practice · fully worked

Original variant: at least one automatically approved packet

A batch contains four independent data packets. For each packet, the checksum-error count is Poisson with mean ln(2), and its latency score is standard normal; these two measurements are independent. A packet is automatically approved only when it has zero checksum errors and a positive latency score. Calculate the probability that at least one packet in the batch is automatically approved.

A
0.2500
B
0.3164
C
0.4219
D
0.6836
E
0.7500

Answer and calculation

One packet is approved with probability exp(−ln)(2) times P(Z>0)=0.5(0.5)=0.25. The complement probability that none of four independent packets is approved is 0.75⁴, so the requested probability is 1-0.75⁴=0.68359375, choice D.

Setup

First calculate the approval probability for one packet.

\Pr(X=0)=e^{-\ln 2}=\frac12

\Pr(Z>0)=\frac12

Model

The two packet-level requirements are independent, and approvals are independent across packets.

q=\Pr(X=0)\Pr(Z>0)=\frac14

A\sim\operatorname{Binomial}\!\left(4,\frac14\right)

Compute

Use the complement of receiving no automatic approvals.

\Pr(A\ge1)=1-\Pr(A=0)

=1-\left(\frac34\right)^4=\frac{175}{256}=0.68359375

Answer

The probability of at least one automatic approval rounds to 0.6836.

\boxed{0.6836\quad\text{(D)}}

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: at least one automatically approved packet

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables