Worked solution · Exam P

SOA Exam P Sample Question #483 Solution

This Exam P sample reference tests Independence and Unions. The given binomial variance and the restriction p≤0.5 imply p=0.10. For one patient the probability of at least one disease is 1-(0.9)²=0.19, so the count variance is 100(0.19)(0.81)=15.39, selecting choice A.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Independence and Unions question

Setup

Let p be the per-patient probability of each disease. The count having disease A is binomial across the 100 independent patients.

\operatorname{Var}(N_A)=100p(1-p)=9

Model

Solve the variance equation and use the stated parameter range to select the smaller root.

p(1-p)=0.09

p^2-p+0.09=0

p\in\{0.10,0.90\}\quad\Longrightarrow\quad p=0.10

Compute

For one patient, use the complement of having neither disease. Then treat the 100 patient-level union indicators as independent Bernoulli trials.

q=\Pr(A\cup B)=1-(1-p)^2=1-(0.90)^2=0.19

N_{A\cup B}\sim\operatorname{Binomial}(100,0.19)

\operatorname{Var}(N_{A\cup B})=100(0.19)(0.81)=15.39

Answer

The variance of the number having at least one disease is 15.39.

\boxed{15.39\quad\text{(A)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

30 original multivariate 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.

BThis uses n²q(1 − q) with the rough union probability q = 0.20, giving 100²(0.20)(0.80) = 1600. A binomial variance contains n, not n².
CThis adds the intersection rather than subtracting it, obtaining q = p + p + p² = 0.21 and then 100(0.21)(0.79) = 16.59. Inclusion-exclusion requires 2p − p² = 0.19.
DWriting the union indicator as I(A) + I(B) − I(A and B), but omitting one of its two covariance corrections, gives per-patient variance 0.09 + 0.09 + 0.0099 − 2(0.009) = 0.1719, hence 17.19. Both covariance terms with the overlap indicator are required.
EThis adds the separate disease-count variances, 9 + 9 = 18, as if the union count were N(A) + N(B). Patients having both diseases are then counted twice.

Original practice · fully worked

Original variant: Poisson visitors using either service

Visitors arrive at an exhibit according to a Poisson process with mean 50 per day. Independently for each visitor, use of an audio guide has probability 0.20 and use of an interactive map has probability 0.30; the two service decisions are independent. Let U be the number of visitors in one day who use at least one of the services. Calculate the variance of U.

A
12.32
B
19.00
C
22.00
D
25.00
E
50.00

Answer and calculation

A visitor uses at least one service with probability 1-(0.8)(0.7)=0.44. Poisson thinning makes U Poisson with mean 50(0.44)=22, and a Poisson variance equals its mean, so choice C is correct.

Setup

First combine the two independent service decisions into one qualifying probability.

q=1-(1-0.20)(1-0.30)=1-(0.80)(0.70)=0.44

Model

Independently retaining each arrival with probability q thins a Poisson count into another Poisson count.

U\sim\operatorname{Poisson}(50q)=\operatorname{Poisson}(22)

Compute

Use equality of the mean and variance for a Poisson random variable.

\operatorname{Var}(U)=50(0.44)=22

Answer

The one-day qualifying-visitor count has variance 22.

\boxed{22.00\quad\text{(C)}}

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How to solve this Independence and Unions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: Poisson visitors using either service

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables