A Defective Bulb and Three Machines: Who Should Be Inspected First?
A bulb-manufacturing unit in Ernakulam runs three machines. Machine A produces 50 per cent of the daily output, machine B 30 per cent and machine C 20 per cent. From long inspection records the unit knows that 2 per cent of A's bulbs, 3 per cent of B's bulbs and 5 per cent of C's bulbs are defective. Bulbs from the three machines are mixed and packed together, so a bulb picked from the finished stock carries no machine label. A consumer complaint arrives about one defective bulb, and the quality manager must decide which machine to inspect first. A supervisor says the answer is obvious: machine C is the worst, since its defect rate is the highest. The manager is not convinced, because machine A makes half of everything. The plant owner wants two things written down: the overall defective rate of the finished stock, and, for a bulb known to be defective, the probability that each machine produced it.
1. Find the overall proportion of defective bulbs in the finished stock.
Read the worked answer
Let A, B, C be the events that a bulb came from that machine, and D the event that it is defective. Then $P(A)=0.50$, $P(B)=0.30$, $P(C)=0.20$, with $P(D|A)=0.02$, $P(D|B)=0.03$, $P(D|C)=0.05$.
By the law of total probability:
$$P(D)=P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)$$ $$=0.50(0.02)+0.30(0.03)+0.20(0.05)=0.010+0.009+0.010=0.029$$
So 2.9% of finished bulbs are defective - about 29 in every 1000. Notice that A and C each contribute 0.010 to the defective pool despite very different defect rates and very different output shares; that coincidence is the heart of the next part.
2. Given that a bulb is defective, find the probability that it came from each machine, and judge the supervisor's claim.
Read the worked answer
Apply Bayes' theorem with the 0.029 found above:
$$P(A|D)=\frac{0.010}{0.029}=0.345,\qquad P(B|D)=\frac{0.009}{0.029}=0.310,\qquad P(C|D)=\frac{0.010}{0.029}=0.345$$
As a check, $0.345+0.310+0.345=1.000$.
Judgement: the supervisor is wrong to single out C. Although C has the highest defect rate (5%), it makes only 20% of the output, so its posterior share (34.5%) merely ties with A, which has a low defect rate but half the output. B is the least likely source (31%).
The posterior combines the prior share with the likelihood: a high defect rate and a large output share pull in opposite directions, and here they exactly cancel for A and C. Inspecting only C first would ignore the machine that is equally implicated. The correct next step is to inspect A and C with equal priority, or to check whether the complaint bulb carries any batch or shift marking.
3. Suppose the defect rate of machine A doubles to 4 per cent while B and C are unchanged. Recompute and comment on what the change shows about base rates.
Read the worked answer
New contribution from A: $0.50\times0.04=0.020$.
$$P(D)=0.020+0.009+0.010=0.039\ (3.9%)$$
$$P(A|D)=\frac{0.020}{0.039}=0.513,\quad P(B|D)=\frac{0.009}{0.039}=0.231,\quad P(C|D)=\frac{0.010}{0.039}=0.256$$
Now machine A is the source of more than half of all defective bulbs, even though its defect rate (4%) is still lower than C's (5%).
What the change shows: the posterior probability of a source depends on the prior share as much as on the defect rate. A moderately bad machine that makes most of the output can dominate the defective pool, while a badly behaved machine that makes little can stay a minor contributor. Any Bayesian argument in a quality report must therefore quote the output shares, not just the defect percentages; quoting 'C is the worst machine' without the base rate is a classic base-rate fallacy.