Defect Attribution and Repair Budget Decision
A factory uses two machines, A and B, to make identical bulbs. Machine A produces 60% of the daily output and machine B produces 40%. Historical records show that 3% of A's bulbs are defective and 5% of B's bulbs are defective. A buyer picks one bulb at random from the mixed daily output and finds it defective. The production manager says: 'Since B has the higher defect rate, B is responsible for most of the defects, so we should spend the limited repair budget on B first.' A quality officer wants the claim checked with conditional probability and with expected defect counts per 10,000 bulbs. The repair that would reduce B's defect rate to 3% costs K150,000; reducing A's defect rate to 2% costs K120,000. Each defective bulb that escapes costs K2,500 in warranty. The daily output is large enough to treat the proportions as probabilities.
1. Calculate the overall probability that a randomly chosen bulb is defective. Hence estimate the expected number of defective bulbs in 10,000 bulbs, and split that expected number by machine.
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P(D)=P(A)P(D|A)+P(B)P(D|B)=0.6(0.03)+0.4(0.05)=0.018+0.020=0.038. In 10,000 bulbs the expected number of defects is 0.038×10000=380. From A: 0.018×10000=180. From B: 0.020×10000=200. So B accounts for slightly more expected defects, but not overwhelmingly more.
2. Given that the chosen bulb is defective, find the probability it came from machine B. Interpret the result in the context of the manager's claim.
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P(B|D)=P(B∩D)/P(D)=0.020/0.038=20/38≈0.5263. About 52.6% of defective bulbs come from B and 47.4% from A. The manager's statement that B is responsible for most defects is technically true in conditional-probability terms, but only by a small margin. It does not by itself show that repairing B is the best use of the budget; the cost of the repair and the size of the defect reduction must also be considered.
3. Evaluate the two repair options using expected warranty cost per 10,000 bulbs. Which repair gives the larger net benefit? State any assumption and a budget caveat.
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Current expected warranty cost is 380×2500=950000. Option B: defect rate 5% to 3%. New P(D)=0.6(0.03)+0.4(0.03)=0.030, so 300 defects. Warranty cost is 300×2500=750000. Saving is 200000; repair cost is 150000; net benefit is 50000. Option A: defect rate 3% to 2%. New P(D)=0.6(0.02)+0.4(0.05)=0.012+0.020=0.032, so 320 defects. Warranty cost is 800000. Saving is 150000; repair cost is 120000; net benefit is 30000. Under these figures B gives the larger net benefit, K50,000 versus K30,000. If the budget is capped at K120,000, B is not affordable and A is the better feasible option. If both repairs can be funded at K270,000, combined P(D)=0.6(0.02)+0.4(0.03)=0.024, giving 240 defects; saving is 140×2500=350000, so net benefit is 350000-270000=80000, better than either single repair.