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
Answer and calculation
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.
0.300(C) 