Verified solution · Exam P

SOA Exam P Sample Question #373 Solution

This Exam P sample reference tests Marginal Distributions. This is a marginal-distribution variance problem for a discrete joint model. Summing over the second coordinate gives masses 3/12, 4/12, and 5/12 at X values 0, 1, and 2. Their first two moments produce variance 23/36, approximately 0.639, so choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Marginal Distributions question

Setup

Only the row sums of the supplied joint mass function are needed. Sum over all three possible values of the second coordinate.

p_X(x)=\sum_{y=0}^{2}\frac{x+y+2}{36}=\frac{3x+9}{36}=\frac{x+3}{12}

p_X(0)=\frac3{12},\qquad p_X(1)=\frac4{12},\qquad p_X(2)=\frac5{12}

Model

Compute the first and second raw moments from the marginal masses.

E[X]=0\left(\frac3{12}\right)+1\left(\frac4{12}\right)+2\left(\frac5{12}\right)=\frac76

E[X^2]=0^2\left(\frac3{12}\right)+1^2\left(\frac4{12}\right)+2^2\left(\frac5{12}\right)=2

Compute

Apply the raw-moment variance identity without rounding the mean.

\operatorname{Var}(X)=E[X^2]-E[X]^2

\operatorname{Var}(X)=2-\left(\frac76\right)^2=\frac{23}{36}=0.6388888889

Answer

The marginal variance rounds to 0.64.

\boxed{0.64\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.

ARounding the mean 7/6 to 1.2 before squaring produces 0.56. The exact mean must be retained through the variance calculation.
CThe value 5/6, or 0.8333, is the factorial moment of X obtained from the mass at 2. It is not the variance.
DThe value 2 is the second raw moment of X; it omits subtraction of the squared mean.
EAdding rather than subtracting the squared mean gives 121/36, or 3.3611, which rounds to 3.36.

Original practice · fully worked

Original variant: variance with a random inspection rate

At the start of a shift, a calibration parameter P is drawn uniformly from 0 to 1. Given its realized value, three inspection lamps act independently and each turns red with probability P. Let N be the number of red lamps. Calculate the variance of N.

A
0.50
B
0.75
C
1.25
D
1.50
E
3.50

Variant answer in brief

The law of total variance separates ordinary binomial variation from shift-to-shift variation in P. Those contributions are 1/2 and 3/4, whose sum is 5/4, so choice C.

Setup

Condition on the calibration parameter so the lamp count has an ordinary binomial model.

N\mid P=p\sim\operatorname{Binomial}(3,p)

E[N\mid P]=3P,\qquad \operatorname{Var}(N\mid P)=3P(1-P)

Model

Apply the law of total variance and use the first two moments of a unit-uniform variable.

\operatorname{Var}(N)=E[\operatorname{Var}(N\mid P)]+\operatorname{Var}(E[N\mid P])

E[P]=\frac12,\qquad E[P^2]=\frac13,\qquad \operatorname{Var}(P)=\frac1{12}

Compute

Evaluate the within-shift and between-shift terms separately, then add them.

E[3P(1-P)]=3\left(\frac12-\frac13\right)=\frac12

\operatorname{Var}(3P)=9\left(\frac1{12}\right)=\frac34

\operatorname{Var}(N)=\frac12+\frac34=\frac54

Answer

The count variance is 1.25.

\boxed{1.25\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #373 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Marginal Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: variance with a random inspection rate

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution