Defect Attribution Across Three Stitching Units
Kitex Uniforms stitches school uniforms in three units and sends everything to one dispatch warehouse. Unit A makes 50% of daily output, unit B 30% and unit C 20%. Last year's quality log gives defect rates of 2% for A, 3% for B and 5% for C; these are treated as exact probabilities, and defects occur independently across uniforms.
One morning the inspector draws a uniform at random from the dispatch lot and finds it defective. The supervisor of unit C protests that the finding says nothing about his unit because C contributes only a fifth of the output. The plant manager wants three numbers: the overall defect rate, the probability that the defective uniform came from C, and the same probability for A, so that the supervisor's argument can be judged on evidence rather than on volume share alone. She also wants to know what happens to both numbers if C's defect rate is halved to 2%.
1. Compute the overall defect rate and the probability that a defective uniform came from unit C.
Read the worked answer
Let $D$ be the event that the drawn uniform is defective. By the law of total probability:
$$P(D)=0.5(0.02)+0.3(0.03)+0.2(0.05)=0.010+0.009+0.010=0.029$$
The overall defect rate is 2.9%, i.e. about 29 defective uniforms per thousand.
By Bayes' theorem:
$$P(C\mid D)=\frac{P(C)P(D\mid C)}{P(D)}=\frac{0.2\times 0.05}{0.029}=\frac{0.010}{0.029}=\frac{10}{29}\approx 0.345$$
So roughly 34.5% of all defects originate in unit C, even though C makes only 20% of the output.
2. Compute $P(A\mid D)$ and use it, together with the defect rates, to judge the supervisor's protest.
Read the worked answer
$$P(A\mid D)=\frac{0.5\times 0.02}{0.029}=\frac{0.010}{0.029}=\frac{10}{29}\approx 0.345$$
$P(A\mid D)=P(C\mid D)$ exactly, because both units contribute the same 0.010 to the overall defect rate: A's low defect rate is cancelled by its large volume, and C's small volume is cancelled by its high defect rate.
The supervisor is therefore wrong on the substance. Being told 'the uniform is defective' makes C as likely as A, a unit five times its size. On a per-uniform basis C is 2.5 times as defect-prone as A (5% against 2%), and C accounts for about a third of all defects from a fifth of the output. His unit is over-represented among defects, not exonerated.
3. If unit C's defect rate is halved to 2%, recompute the overall defect rate and $P(C\mid D)$, and interpret the change.
Read the worked answer
$$P(D)=0.5(0.02)+0.3(0.03)+0.2(0.02)=0.010+0.009+0.004=0.023$$
Overall defect rate falls from 2.9% to 2.3%, a relative reduction of about 20.7% - roughly one fifth of all defects removed by fixing one unit.
$$P(C\mid D)=\frac{0.004}{0.023}=\frac{4}{23}\approx 0.174$$
C's share of the defects drops from about 34.5% to about 17.4%, now slightly below its 20% share of output, so C would no longer be over-represented. Note the arithmetic of attribution: halving C's rate cut its own contribution from 0.010 to 0.004, but the total also shrank, so the ratio moves more than the raw numbers suggest.