Worked solution · Exam P
This Exam P sample reference tests Bayes' Theorem. A positive indication has defective-board mass 0.10(0.80)=0.08 and nondefective-board mass 0.90(0.60)=0.54. Bayes' formula gives 0.08/0.62=4/31, so choice D is correct.
Independent solution
Setup
Let D denote a defective item and T a positive test indication. Convert the stated accuracy on nondefective items into a false-positive probability.
Model
Partition the positive-test event by the item's true condition and apply Bayes' formula.
Compute
Insert the prior and the two conditional indication probabilities.
Answer
The posterior defective probability is 4/31.
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Original practice · fully worked
One percent of data packets entering a network are corrupted. Two monitoring rules evaluate each packet. Conditional on whether a packet is corrupted, the rules act independently. Each rule flags a corrupted packet with probability 0.90 and flags an intact packet with probability 0.05. A packet triggers both rules. Calculate the probability that it is corrupted.
Conditional independence makes the two-flag likelihoods 0.90² for corrupted packets and 0.05² for intact packets. Bayes' formula gives 0.0081/(0.0081+0.002475)=0.76596, so choice D is correct.
Setup
Let C denote corruption and F the event that both rules flag the packet.
Model
Conditional independence lets each two-rule likelihood be written as a product.
Compute
Weight both likelihoods by their prior probabilities and normalize the corrupted-packet mass.
Answer
A packet triggering both rules is corrupted with probability about 0.7660.
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