Verified solution · Exam P

SOA Exam P Sample Question #244 Solution

This Exam P sample reference tests Joint Distributions. Summing the joint table over the other variable gives probabilities 0.915, 0.072, 0.012, and 0.001 for hospitalization counts 0 through 3. Their weighted sum is 0.099, choice B.

Independent analysis by ActuaryProof · reviewed August 4, 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 Joint Distributions question

Setup

First reduce the joint probability table to the marginal distribution of the hospitalization count Y.

p_Y(y)=\sum_x p_{X,Y}(x,y)

Model

Add each table column associated with the same value of Y.

p_Y(0)=0.915,\quad p_Y(1)=0.072,\quad p_Y(2)=0.012,\quad p_Y(3)=0.001

Compute

Weight the possible counts by their marginal probabilities.

\operatorname{E}[Y]=0(0.915)+1(0.072)+2(0.012)+3(0.001)

\operatorname{E}[Y]=0.072+0.024+0.003=0.099

Answer

The expected annual hospitalization count is 0.099.

\boxed{0.099\quad\text{(B)}}

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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 0.085 is P(Y at least 1)=0.072+0.012+0.001. It counts whether any hospitalization occurs but does not weight outcomes with two or three events.
CThe value 0.410 is the expectation of the table's other count variable. It results from summing in the wrong direction and answering for X rather than Y.
DThe value 1.000 is the sum of all probabilities in the joint table. That checks normalization but is not an expected count.
EThe value 1.500 is the mean of 0, 1, 2, and 3 under an unstated uniform distribution. The table probabilities are highly nonuniform.

Original practice · fully worked

Original variant: expected inspections across routing classes

A shipment is assigned to one of three routes. Route A is used with probability 0.20 and requires no inspection. Route B is used with probability 0.50; there the shipment receives one inspection with probability 0.60 and two with probability 0.40. The remaining route probability belongs to C, where the inspection count is Poisson with mean 3. Calculate the unconditional expected number of inspections.

A
0.7000
B
0.9000
C
1.4000
D
1.4667
E
1.6000

Variant answer in brief

The conditional means are 0, 1.4, and 3 for routes A, B, and C. Weighting them by the routing probabilities gives 0.20(0)+0.50(1.4)+0.30(3)=1.6, choice E.

Setup

Let R denote the selected route and N the number of inspections.

\left(\Pr(R=A),\Pr(R=B),\Pr(R=C)\right)=(0.20,0.50,0.30)

Model

Find the expected count within each route.

\operatorname{E}[N\mid R=A]=0

\operatorname{E}[N\mid R=B]=1(0.60)+2(0.40)=1.4

\operatorname{E}[N\mid R=C]=3

Compute

Apply the law of total expectation with the actual route weights.

\operatorname{E}[N]=\sum_r\Pr(R=r)\operatorname{E}[N\mid R=r]

\operatorname{E}[N]=0.20(0)+0.50(1.4)+0.30(3)=1.6

Answer

Across all shipments, the mean inspection count is 1.6.

\boxed{1.6000\quad\text{(E)}}

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ActuaryProof. “SOA Exam P Sample Question #244 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Joint Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: expected inspections across routing classes

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution