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.
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
Variant answer in brief
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
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
Model
Independently retaining each arrival with probability q thins a Poisson count into another Poisson count.
U∼Poisson(50q)=Poisson(22)
Compute
Compute
Use equality of the mean and variance for a Poisson random variable.
Var(U)=50(0.44)=22
Answer
Answer
The one-day qualifying-visitor count has variance 22.
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