Tracing defective bolts across three machines
A fastener unit operates three machines, M1, M2 and M3. On a typical day M1 makes 50% of the output, M2 30% and M3 20%. Historical inspection records show defect rates of 2% for M1, 3% for M2 and 5% for M3. One day's total output is 12,000 bolts. An inspector picks one bolt at random from the day's output and finds it defective. The production manager says the defective bolt is most likely from M3 because M3 has the highest defect rate. The quality team wants the overall defective rate, the expected number of defectives in the day's output, and the probability that the picked defective bolt came from each machine before deciding where to focus an improvement drive.
1. Find the overall defective rate and the expected number of defective bolts in the day's output.
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Use the law of total probability:
P(defective) = P(M1)P(D|M1) + P(M2)P(D|M2) + P(M3)P(D|M3)
= (0.50)(0.02) + (0.30)(0.03) + (0.20)(0.05)
= 0.010 + 0.009 + 0.010
= 0.029.
The overall defective rate is 2.9%. For 12,000 bolts, the expected number of defectives is 12,000 times 0.029 = 348 bolts.
2. Find the probability that the defective bolt came from each machine. Is the manager's claim that it is most likely from M3 correct?
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Apply Bayes' rule. The joint probabilities are:
P(M1 and D) = 0.010, P(M2 and D) = 0.009, P(M3 and D) = 0.010.
Since P(D) = 0.029:
P(M1|D) = 0.010 / 0.029 = 0.3448, or 34.48%.
P(M2|D) = 0.009 / 0.029 = 0.3103, or 31.03%.
P(M3|D) = 0.010 / 0.029 = 0.3448, or 34.48%.
The manager's claim is not correct as stated. M3 has the highest defect rate, but M1 produces far more bolts. M1 and M3 are tied at about 34.5%, and M2 is close behind. The improvement drive should consider both the defect rate and the production volume; M1 and M3 deserve attention.
3. A separate lot contains 20 bolts from M1, of which 3 are defective. If 5 bolts are inspected without replacement, find the probability that exactly one is defective and explain why a binomial model is not appropriate.
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Use the hypergeometric distribution:
P(exactly one defective) = C(3,1) times C(17,4) / C(20,5).
C(3,1) = 3; C(17,4) = 2,380; C(20,5) = 15,504.
Probability = (3 times 2,380) / 15,504 = 7,140 / 15,504 = 0.4605 approximately.
A binomial model would assume independent trials with a constant probability of defect on every draw. Here the bolts are drawn without replacement, so after each draw the remaining lot has a different proportion of defectives. The hypergeometric model accounts for this changing composition.