Defect Rate and the Wrong Conditional
A factory has two assembly lines. Line A makes 60% of the daily output and Line B makes 40%. Historical records show that 2% of Line A's items are defective and 5% of Line B's items are defective. An inspector picks one finished item at random and finds it defective. The line supervisor says: 'Since Line A produces most of our output, a defective item is more likely to have come from Line A than from Line B.' The plant manager wants the posterior probabilities, the overall defect rate, and a decision on a proposal that moves 10 percentage points of total output from B to A while each line's defect rate stays unchanged.
1. Compute the overall defect rate and P(A|defective), P(B|defective). Is the supervisor right?
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P(D)=P(A)P(D|A)+P(B)P(D|B)=0.6(0.02)+0.4(0.05)=0.012+0.020=0.032, so the overall defect rate is 3.2%.
Then P(A|D)=0.012/0.032=0.375 and P(B|D)=0.020/0.032=0.625.
The supervisor is wrong: although Line A makes more items, a defective item is more likely to come from Line B, because B's defect rate is much higher.
2. Under the proposal, A makes 70% and B makes 30%. Compute the new overall defect rate and compare. Does the proposal reduce defects?
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Under the proposal, A makes 70% and B makes 30%. New P(D)=0.7(0.02)+0.3(0.05)=0.014+0.015=0.029, or 2.9%.
This is lower than 3.2% by 0.3 percentage points, so the proposal reduces the overall defect rate, assuming the stated defect rates stay unchanged.
3. Are line and defect status independent? Use a numerical comparison.
Read the worked answer
Line and defect status are not independent. For independence we would need P(A|D)=P(A). But P(A|D)=0.375 while P(A)=0.6.
Equivalently, P(A∩D)=0.012, whereas P(A)P(D)=0.6(0.032)=0.0192. Since these differ, defect status carries information about the line.