Defect Source and Batch Acceptance in a Bolt Factory
A factory uses three machines A, B and C to make bolts. Machine A produces 50% of daily output, B 30%, and C 20%. Historical records show defect rates of 2% for A, 3% for B and 5% for C. A quality inspector selects one bolt at random and finds it defective. The production manager wants to know which machine most likely produced it, because maintenance will be scheduled on that machine. Separately, a customer will accept a batch of 20 bolts only if at most one is defective. The inspector plans to use the overall defect rate to estimate the acceptance probability, but a colleague warns that defects may cluster because bolts from the same machine are packed together. The manager must combine the machine shares with the defect rates and assess the batch acceptance calculation.
1. Find the overall probability that a randomly selected bolt is defective, and the probability that a defective bolt came from machine B. Compare with the other machines.
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P(D)=0.50×0.02+0.30×0.03+0.20×0.05=0.010+0.009+0.010=0.029. By Bayes, P(B|D)=0.009/0.029=9/29≈0.310. Also P(A|D)=0.010/0.029=10/29≈0.345 and P(C|D)=0.010/0.029=10/29≈0.345. So B is not the most likely source; A and C are tied and each slightly more likely than B. Scheduling maintenance on B alone is not supported by this evidence.
2. Using the overall defect rate, estimate the probability that a batch of 20 bolts contains at most one defective. State the assumptions.
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Let X be the number of defectives, modelled as Binomial(n=20, p=0.029). Then P(X≤1)=P(X=0)+P(X=1)=0.971^20 + 20(0.029)(0.971)^19. Now 0.971^20≈0.555 and 0.971^19≈0.572, so the second term is about 20×0.029×0.572≈0.332. The total is about 0.887. This assumes independent defects with the same probability 0.029 for every bolt.
3. The colleague warns that defects may cluster by machine. Explain how this affects the calculation and what the inspector should do.
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Clustering means the binomial independence assumption fails: a batch packed from one machine may have a defect rate of 2%, 3% or 5%, not 2.9%. The pooled binomial calculation then misstates the acceptance probability and understates variability. The inspector should record the machine source, compute acceptance separately for each machine rate, or use a mixture model. For example, at p=0.05, P(X≤1)=0.95^20+20(0.05)(0.95)^19≈0.358+0.377=0.735, noticeably lower than 0.887.