Screening Test on Two Soldering Lines
A circuit-board plant runs two soldering lines. Line A makes 60% of daily output and has a 2% defect rate; line B makes 40% and has a 5% defect rate. Every board is screened by an automated tester. The tester flags a defective board with probability 0.90, but it also flags 4% of good boards as false alarms. A flagged board is routed to rework, which costs ₹180 per board; a missed defective board that escapes costs ₹1,200 in field repair. The plant manager sees that most flagged boards come from line A and concludes that line A is the main quality problem. The quality engineer is asked to compute the overall defect rate, the probability that a flagged board is truly defective, and the probability that a truly defective board came from line B. The engineer must also advise whether the manager's conclusion is supported.
1. Compute the overall defect rate and the probability that a defective board came from line B.
Read the worked answer
Let D be 'defective', A be 'from line A', and B be 'from line B'.
P(D) = P(A)P(D|A) + P(B)P(D|B) = (0.60)(0.02) + (0.40)(0.05) = 0.012 + 0.020 = 0.032.
The overall defect rate is 3.2%.
P(B|D) = P(B)P(D|B)/P(D) = (0.40)(0.05)/0.032 = 0.020/0.032 = 0.625.
So 62.5% of defective boards come from line B, even though line B makes only 40% of output. Line B has the higher defect rate and contributes more defective boards.
2. Compute the probability that a flagged board is truly defective, using the tester's sensitivity and false-alarm rate.
Read the worked answer
Let Pos be 'flagged by the tester'. We are given P(Pos|D) = 0.90 and P(Pos|not D) = 0.04.
First find P(Pos): P(Pos) = P(Pos|D)P(D) + P(Pos|not D)P(not D) = (0.90)(0.032) + (0.04)(0.968) = 0.0288 + 0.03872 = 0.06752.
Then by Bayes: P(D|Pos) = P(Pos|D)P(D)/P(Pos) = 0.0288/0.06752 ≈ 0.4265.
Only about 42.7% of flagged boards are truly defective. The false-alarm rate of 4% on a large good population creates more flags than the 90% detection of a 3.2% defect population.
3. The manager says most flagged boards come from line A, so line A is the main quality problem. Test this claim numerically and advise.
Read the worked answer
Compute flagged boards by line:
P(A and Pos) = 0.60[(0.02)(0.90) + (0.98)(0.04)] = 0.60(0.018 + 0.0392) = 0.60(0.0572) = 0.03432.
P(B and Pos) = 0.40[(0.05)(0.90) + (0.95)(0.04)] = 0.40(0.045 + 0.038) = 0.40(0.083) = 0.03320.
Total P(Pos) = 0.06752, so P(A|Pos) = 0.03432/0.06752 ≈ 0.508, P(B|Pos) = 0.03320/0.06752 ≈ 0.492.
Flagged boards are almost evenly split, with only a slight excess from A. That is not strong evidence that A is the main quality problem. On defect contribution, B contributes 0.020 of the 0.032 overall defect rate, while A contributes 0.012. The better conclusion is that B has the worse defect rate and is the larger source of true defects; the manager's inference from raw flag counts confuses volume with defect propensity.