This Exam P sample reference tests Bayes' Theorem. Bayes' theorem combines the 20% prior class share with the two class-specific Poisson likelihoods for a count of two. The posterior probability is 0.170298, which matches choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.20(exp(−1)/2)=0.036788 is the joint numerator P(A and N=2), not the posterior after division by the mixture likelihood.
CThe value exp(−1)/2=0.183940 is the within-class likelihood of two events and omits both the prior and Bayes normalization.
DThe value 0.200 is the original class proportion; it ignores the information supplied by the observed count.
EThe value (exp(−1)/2)/(exp(−1)/2+9 exp(−3)/2)=0.450853 treats the two classes as equally likely before observing the count.
Original practice · fully worked
Original variant: sensor warning classification
A warehouse uses two sensor models. Model R makes up 30% of installed sensors and produces a Poisson number of warning signals per week with mean 0.5. Model S makes up the remainder and has mean 2. Given that a randomly selected sensor produced exactly one warning last week, calculate the probability that it is model R.
A 0.091
B 0.300
C 0.303
D 0.324
E 0.528
Variant answer in brief
Weighting the two one-warning likelihoods by the 30% and 70% model shares gives a posterior model-R probability of 0.324407, so choice D is correct.
Setup
Setup
Let W be the weekly warning count and write the prior model shares and conditional means.
Pr(R)=0.30,Pr(S)=0.70
W∣R∼Poisson(0.5),W∣S∼Poisson(2)
Model
Model
Use the one-warning mass under each model as the likelihood in Bayes' formula.
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