SOA Exam P Sample Question #606. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #606 — Verified Solution

Independent solution

How to solve this Joint Distributions question

Answer in brief

Fixing the second count at one gives relative masses 12, 20, and 8 for first-count values zero, one, and two. After normalization their mean is 0.9 and second moment is 1.3, so the conditional variance is 1.3-0.9^2=0.49 and choice B.

Setup

Evaluate the joint probability rule at the conditioning value. Common factors cancel, leaving three relative masses.

p(0,1):p(1,1):p(2,1)=12:20:8

Model

Normalize the three masses by their total of forty to obtain the conditional distribution.

\bigl(\Pr(X=0\mid Y=1),\Pr(X=1\mid Y=1),\Pr(X=2\mid Y=1)\bigr)=\left(\frac{12}{40},\frac{20}{40},\frac8{40}\right)

\bigl(\Pr(X=0\mid Y=1),\Pr(X=1\mid Y=1),\Pr(X=2\mid Y=1)\bigr)=(0.3,0.5,0.2)

Compute

Calculate the conditional first and second moments, then subtract the squared mean.

\operatorname{E}[X\mid Y=1]=0(0.3)+1(0.5)+2(0.2)=0.9

\operatorname{E}[X^2\mid Y=1]=0^2(0.3)+1^2(0.5)+2^2(0.2)=1.3

\operatorname{Var}(X\mid Y=1)=1.3-(0.9)^2=0.49

Answer

The conditional variance is 49/100.

\boxed{\frac{49}{100}\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

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

AThis can result from losing a conditional mass or using an incorrect normalizing total before calculating moments. The three retained weights must sum to forty.
CThis is the variance of a discrete uniform variable on zero, one, and two. The actual conditional masses are not equal.
DThe value 0.70 is the conditional standard deviation, the square root of 0.49, rather than the variance.
EThis treats a rounded moment or scale value as the variance. Variance requires the second moment minus the square of the full conditional mean.

Original practice · fully worked

Original variant: horizontal error on a disk chord

A robotic alignment error (X,Y) is uniformly distributed over the unit disk x^2+y^2<1. On a particular run, the vertical error is observed to be Y=3/5. Calculate the conditional variance of the horizontal error X.

A
4/75
B
16/75
C
1/4
D
16/25
E
4/5

Variant answer in brief

At vertical coordinate 3/5, the disk's horizontal chord runs from -4/5 to 4/5. The conditional horizontal coordinate is uniform on that chord, whose variance is (8/5)^2/12=16/75, selecting choice B.

Setup

Intersect the unit disk with the observed vertical coordinate to find the allowable horizontal range.

x^2+\left(\frac35\right)^2<1

|x|<\sqrt{1-\frac9{25}}=\frac45

Model

The joint density is constant over the disk. Along a fixed horizontal chord, the conditional density is therefore uniform.

X\mid\left(Y=\frac35\right)\sim\operatorname{Uniform}\!\left(-\frac45,\frac45\right)

Compute

Use the uniform variance formula with full chord width eight-fifths.

\operatorname{Var}\!\left(X\mid Y=\frac35\right)=\frac{\left(\frac85\right)^2}{12}

\operatorname{Var}\!\left(X\mid Y=\frac35\right)=\frac{16}{75}=0.213333\ldots

Answer

The conditional variance is 16/75.

\boxed{\frac{16}{75}\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #606 — Verified Solution.” Verified 2026-07-30.

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