Worked solution · Exam P

SOA Exam P Sample Question #569 Solution

This Exam P sample reference tests Sums of Independent Random Variables. A total below two consists of no losses, one hail loss, or one theft loss. The three independent-event probabilities sum to 0.1726357, so choice A is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 Sums of Independent Random Variables question

Setup

Let H and T be the ten-year counts for the two independent loss types.

H\sim\operatorname{Binomial}(10,0.2)

T\sim\operatorname{Binomial}(10,0.1)

Model

The total is below two in exactly three disjoint count combinations.

\Pr(H+T<2)=\Pr(H=0,T=0)+\Pr(H=1,T=0)+\Pr(H=0,T=1)

Compute

Use independence to multiply the component probabilities within each combination.

\Pr(H=0,T=0)=0.8^{10}0.9^{10}=0.0374390624

\Pr(H=1,T=0)=10(0.2)0.8^9 0.9^{10}=0.0935976561

\Pr(H=0,T=1)=0.8^{10}10(0.1)0.9^9=0.0415989583

\Pr(H+T<2)=0.1726356767

Answer

The probability rounds to 0.17.

\boxed{0.17\quad\text{(A)}}

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.

BThis collapses each year to an indicator for exactly one loss, with probability 0.2(0.9)+0.1(0.8)=0.26, and then computes P(Binomial(10,0.26)≤1)=0.222245, losing years with two losses.
CThis uses the same incorrect one-loss-per-year model as choice B and then takes its complement, 1-0.222245=0.777755.
DThis approximates the total by Poisson(3) and then takes the complement of the desired small-count event: 1-4 exp(−3)=0.800852.
EThis takes the complement of the exact result, 1-0.1726357=0.8273643, thereby answering for at least two losses.

Original practice · fully worked

Original variant: infer an alarm rate

Two alarm channels are observed for six shifts. Channel A triggers independently on each shift with probability 0.10. Channel B is independent of A and across shifts, but its trigger probability q is unknown. The probability that neither channel triggers during all six shifts is 0.0945851. Calculate the expected total number of triggers from both channels over the six shifts.

A
0.25
B
0.35
C
0.60
D
1.50
E
2.10

Answer and calculation

The no-trigger probability is (0.9)⁶(1-q)⁶=0.0945851, so 0.9(1-q)=0.675 and q=0.25. Linearity gives expected total 6(0.10+0.25)=2.10, selecting E.

Setup

Write the six-shift probability of no trigger from either independent channel.

0.0945850803=0.9^6(1-q)^6

Model

Take the sixth root to recover the per-shift no-trigger factor.

0.9(1-q)=0.0945850803^{1/6}=0.675

Compute

Solve for q and add the expected trigger counts from the two channels.

q=0.25

E[N_A+N_B]=6(0.10)+6(0.25)=2.10

Answer

The two channels produce 2.10 triggers on average.

\boxed{2.10\quad\text{(E)}}

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How to solve this Sums of Independent Random Variables question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: infer an alarm rate

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P Multivariate Random Variables