Two-Machine Defect Screening and Base Rates
A factory uses two machines, A and B, to make identical components. Machine A produces 60% of the daily output and has a defect rate of 3%. Machine B produces the remaining 40% and has a defect rate of 5%. At the end of a day, a quality inspector picks one component at random and finds it defective. The production manager says: 'Machine B is clearly the problem. It has the higher defect rate, so most defective components must come from B; we should service B first.' The inspector is not convinced and wants a numerical check. A second component is then picked at random from the day's output, independently of the first. The factory also wants to know the probability that exactly one of the two inspected components is defective, and whether the two events 'first is defective' and 'second is defective' are independent.
1. Calculate the probability that a randomly chosen component is defective, and the probability that a defective component came from Machine B. Test the manager's claim.
Read the worked answer
Let A = made by A, B = made by B, D = defective. P(A)=0.6, P(B)=0.4, P(D|A)=0.03, P(D|B)=0.05.
By the law of total probability, P(D)=0.6×0.03+0.4×0.05=0.018+0.020=0.038.
Then P(B|D)=P(B and D)/P(D)=0.020/0.038=0.5263 (52.6%).
So about 52.6% of defective components come from B and 47.4% from A. The manager is correct only narrowly: B supplies slightly more than half of the defectives, but this is not caused by the defect rate alone. Per 1000 components, A contributes 600×0.03=18 defects and B contributes 400×0.05=20 defects. The difference is only 2 defects per 1000, so the higher defect rate alone does not justify treating B as the sole problem.
2. Find the probability that exactly one of two independently chosen components is defective, and state whether the events 'first is defective' and 'second is defective' are independent.
Read the worked answer
Let D1 and D2 be the events that the first and second components are defective. The components are chosen independently, so P(D1)=P(D2)=0.038 and P(not D)=0.962.
Exactly one defective occurs as (D1 and not D2) or (not D1 and D2). These are mutually exclusive, so P(exactly one)=2×0.038×0.962=0.073112.
Independence holds because the selection is stated to be independent: P(D1 and D2)=0.038×0.038=0.001444=P(D1)P(D2). Equivalently, knowing the first result does not change the probability for the second.
3. If the factory can reduce one machine's defect rate by 1 percentage point, which machine should it choose to reduce the expected number of defects per 1000 components? Show calculations.
Read the worked answer
Current defects per 1000: A gives 600×0.03=18; B gives 400×0.05=20; total 38.
Reduce A from 3% to 2%: A defects become 600×0.02=12, saving 6 per 1000; total 32. Reduce B from 5% to 4%: B defects become 400×0.04=16, saving 4 per 1000; total 34.
Reducing A's rate by 1 percentage point saves more expected defects (6 vs 4), because A has the larger output share. The manager's focus on the higher defect rate alone is incomplete: base rates matter. If the goal is expected defect count, A is the better target; if the goal is risk per component from B, B may still be prioritised.