This Exam P sample reference tests Bayes Theorem. This problem updates a persistent two-class risk mixture after two claim-free observations. Dividing the four-period joint probability by the observed two-period probability gives 0.957093, so choice E.
Let H and L denote the two fixed risk classes. Let A be the observed pair of event-free periods and B the corresponding event for the next two periods.
Pr(H)=0.10,Pr(L)=0.90
Pr(A∣H)=0.802,Pr(A∣L)=0.992
Model
Model
Conditional independence across periods gives four-period probabilities within each class. Average them using the class priors.
Pr(A∩B)=0.10(0.804)+0.90(0.994)
Pr(A)=0.10(0.802)+0.90(0.992)
Compute
Compute
Form the conditional ratio of the joint four-period event to the observed two-period event.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.10(0.80⁴)+0.90(0.99⁴)=0.905496 is the joint probability of four event-free periods before conditioning on the first two.
BThe value 0.90(0.99²)/[0.10(0.80²)+0.90(0.99²)]=0.932353 is the posterior probability of the lower-risk class, not the probability of two future event-free periods.
CThe value 0.10(0.80²)+0.90(0.99²)=0.946090 is the unconditional probability for any two periods and ignores the information in the observed pair.
DSquaring the one-period posterior predictive probability gives 0.954816. Future periods are independent within a known class, but remain positively dependent after the persistent class is averaged out.
Original practice · fully worked
Original variant: posterior prediction for a new batch
A batch has an unknown defect probability Theta with a Beta(2,3) prior distribution. Five inspected items from the batch contain exactly one defective item. Conditional on these observations, calculate the probability that the next inspected item is defective.
A 0.200
B 0.250
C 0.300
D 0.400
E 0.700
Variant answer in brief
One observed defect and four nondefects update the Beta(2,3) prior to Beta(3,7). The posterior predictive defect probability is the posterior mean 3/(3+7)=0.300, so choice C.
Setup
Setup
Write the prior kernel and the likelihood contributed by one defect and four nondefects.
π(θ)∝θ2−1(1−θ)3−1
L(θ)∝θ(1−θ)4
Model
Model
Multiplying the two kernels identifies the posterior beta parameters.
π(θ∣data)∝θ2(1−θ)6
Θ∣data∼Beta(3,7)
Compute
Compute
The next Bernoulli outcome has posterior predictive probability equal to the posterior mean of Theta.
Pr(next defective∣data)=E[Θ∣data]
=3+73=0.300
Answer
Answer
The posterior predictive defect probability is 0.300.
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