Worked solution · Exam P

SOA Exam P Sample Question #494 Solution

This Exam P sample reference tests Binomial Distribution. The count is binomial. Its mean fixes the common success probability at 3/4, after which np(1-p)=12(3/4)(1/4)=2.25, so choice C is correct.

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 Binomial Distribution question

Setup

Represent each person's qualifying outcome by an independent Bernoulli indicator and let N be the sum of the twelve indicators.

N\sim\operatorname{Binomial}(12,p)

\operatorname{E}[N]=12p=9

Model

First recover the Bernoulli probability from the supplied mean, then use the binomial variance formula.

p=\frac{9}{12}=\frac{3}{4}

\operatorname{Var}(N)=np(1-p)

Compute

Substitute n=12, p=3/4, and 1-p=1/4.

\operatorname{Var}(N)=12\left(\frac{3}{4}\right)\left(\frac{1}{4}\right)

\operatorname{Var}(N)=\frac{9}{4}=2.25

Answer

The variance of the count is 2.25.

\boxed{2.25\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate 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.

AThis reports the recovered individual probability p=9/12=0.75 rather than the variance of the twelve-person count.
BThis is the standard deviation, √2.25 = 1.50. The question asks for variance, so the square root should not be taken.
DThis repeats the supplied mean np=9.00. A binomial variance includes the additional factor 1-p=1/4.
EThis reports the number of trials n=12.00 and ignores both the success and failure probabilities in np(1-p).

Original practice · fully worked

Original variant: infer a dashboard's expected pass count

A dashboard totals the pass indicators from 16 independent devices, all with the same pass probability p. The variance of the displayed total is 3.84, and engineers know p is greater than 0.50. Calculate the expected number of passing devices.

A
3.84
B
6.40
C
8.00
D
9.60
E
12.16

Answer and calculation

The variance equation gives p(1-p)=0.24, whose roots are 0.40 and 0.60. The condition p>0.50 selects 0.60, so the expected count is 16(0.60)=9.60 and choice D.

Setup

Let T be the total number of passing devices. Equal independent pass probabilities make T binomial.

T\sim\operatorname{Binomial}(16,p)

\operatorname{Var}(T)=16p(1-p)=3.84

Model

Solve the variance equation for its two symmetric probability roots and use the directional condition to choose one.

p(1-p)=0.24

p^2-p+0.24=0

Compute

The quadratic roots are 0.40 and 0.60; only the larger root satisfies the engineering information.

p=\frac{1\pm\sqrt{1-0.96}}{2}\in\{0.40,0.60\}

\operatorname{E}[T]=16(0.60)=9.60

Answer

The expected number of passing devices is 9.60.

\boxed{9.60\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: infer a dashboard's expected pass count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables