Worked solution · Exam P

SOA Exam P Sample Question #323 Solution

This Exam P sample reference tests Independence. This is an upper tail of a sum of four independent, nonidentical Bernoulli indicators. The exactly-three cases total 0.3006 and the all-four case contributes 0.0891, giving 0.3897 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Independence question

Setup

Represent each possible sale by an independent indicator and let N count the successful indicators.

(p_1,p_2,p_3,p_4)=(0.45,0.55,0.60,0.60)

N=I_1+I_2+I_3+I_4

Model

More than two successes means exactly three or exactly four. For exactly three, enumerate the four choices of the single failure.

\Pr(N=3)=\sum_{j=1}^{4}(1-p_j)\prod_{i\ne j}p_i

Compute

Evaluate the disjoint three-success cases and add the all-success case.

\Pr(N=3)=0.1089+0.0729+0.0594+0.0594=0.3006

\Pr(N=4)=(0.45)(0.55)(0.60)(0.60)=0.0891

\Pr(N>2)=0.3006+0.0891=0.3897

Answer

The probability rounds to 0.39.

\boxed{0.39\quad\text{(C)}}

Test the method on unseen problems

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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 0.24 rounds P(N at most 1)=0.0396+0.1996=0.2392. That lower tail allows too few successes rather than more than two.
BThe value 0.30 is the probability of exactly three successes, rounded from 0.3006. It omits the 0.0891 probability that all four succeed.
DThe value 0.61 is the complementary probability P(N at most 2)=1-0.3897=0.6103.
EThe value 0.76 rounds P(N at least 2)=0.3711+0.3006+0.0891=0.7608. It incorrectly includes the exactly-two outcome.

Original practice · fully worked

Original variant: quadratic packet-processing charge

Five remote stations independently decide whether to upload a calibration packet during a service window. Their upload probabilities are 0.10, 0.30, 0.50, 0.70, and 0.90. If N packets arrive, the processing charge is 5N squared credits. Calculate the expected processing charge.

A
4.25 credits
B
12.50 credits
C
31.25 credits
D
35.50 credits
E
125.00 credits

Answer and calculation

The packet count has mean 2.5 and variance 0.85 because it is a sum of independent Bernoulli indicators. Thus E[N squared]=0.85+2.5 squared=7.10, making the expected charge 5(7.10)=35.50 credits and choice D.

Setup

Write the packet count as a sum of five independent upload indicators.

N=I_1+\cdots+I_5

(p_1,\ldots,p_5)=(0.10,0.30,0.50,0.70,0.90)

Model

Add Bernoulli means and variances, then convert the variance to a second moment.

E[N]=\sum_i p_i=2.50

\operatorname{Var}(N)=\sum_i p_i(1-p_i)=0.85

E[N^2]=\operatorname{Var}(N)+E[N]^2

Compute

Evaluate the second moment and apply the quadratic charge rule.

E[N^2]=0.85+(2.50)^2=7.10

E[5N^2]=5(7.10)=35.50

Answer

The expected processing charge is 35.50 credits.

\boxed{35.50\text{ credits}\quad\text{(D)}}

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How to solve this Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: quadratic packet-processing charge

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability