Verified solution · Exam P

SOA Exam P Sample Question #83 Solution

This Exam P sample reference tests Conditional Distributions. Normalize the row for zero tornadoes in county P, then compute the first two moments of county Q. The conditional variance is 0.9856, choice D.

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 Conditional Distributions question

Setup

Restrict the joint table to the row where the first county has zero tornadoes. The row probabilities sum to 0.25.

P(P=0)=0.12+0.06+0.05+0.02=0.25

Model

Normalize the four cells in that row to obtain the conditional distribution of the second county's count.

P(Q=0,1,2,3\mid P=0)=\frac1{25}(12,6,5,2)

Compute

The conditional mean is 0.88 and the second raw moment is 1.76. Subtracting the squared mean gives variance 0.9856.

E[Q\mid P=0]=\frac{22}{25}

E[Q^2\mid P=0]=\frac{44}{25}

\operatorname{Var}(Q\mid P=0)=\frac{44}{25}-\left(\frac{22}{25}\right)^2=0.9856

Answer

The conditional variance is approximately 0.99, selecting choice D.

\boxed{0.99\quad\text{(D)}}

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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.

AChoice A uses the unnormalized joint-row probabilities as though they were conditional probabilities.
BChoice B omits one support value from the conditional distribution.
CChoice C rounds the conditional mean before squaring it.
EChoice E reports the conditional second raw moment without subtracting the squared mean.

Original practice · fully worked

Original variant: conditional queue-length variation

The joint probabilities for server state S=0 and queue length Q=0,1,2,3 are 0.08, 0.12, 0.16, and 0.04. Given S=0, find Var(Q).

A
0.80
D
0.84
C
1.00
B
1.14
E
1.25

Variant answer in brief

The row total is 0.40, giving conditional probabilities 0.2,0.3,0.4,0.1. Their mean is 1.4 and second moment 2.8, so variance is 0.84.

Setup

Restrict the joint table to the row where the server state is zero. The four row probabilities sum to 0.40.

P(S=0)=0.40

P(Q=0,1,2,3\mid S=0)=(0.2,0.3,0.4,0.1)

Model

Normalize that row to obtain conditional queue-length probabilities 0.20, 0.30, 0.40, and 0.10.

E[Q\mid S=0]=1.4

E[Q^2\mid S=0]=0.3+1.6+0.9=2.8

Compute

The conditional mean is 1.4 and second raw moment is 2.8, giving variance 0.84.

\operatorname{Var}(Q\mid S=0)=2.8-1.4^2=0.84

Answer

The conditional queue-length variance is 0.84, selecting choice D.

\boxed{0.84\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #83 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Conditional Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: conditional queue-length variation

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution