Exercise 13.3 · Q11: Three operators: defective item produced by A, $P = \frac{5}{34}$

Basics you need

Solution

1. Name the events

   Let

   • $E_1$⁠: the item is produced by operator A;
   • $E_2$⁠: the item is produced by operator B;
   • $E_3$⁠: the item is produced by operator C;
   • $D$⁠: the item is defective.

   $E_1$⁠, $E_2$⁠, $E_3$ form a partition of the sample space (the items produced): each item is produced by exactly one operator, and each operator produces some items.

   The operator is the cause; "defective" is the result we observe. We write $D$ for the result because A, B and C already name the operators.
   Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem

2. The six probabilities as fractions

   The question gives each operator's share of the time on the job. Reading that as each operator's share of the items produced (the operators work at the same rate), the chance that an item was produced by each operator is

   $$\begin{aligned} P(E_1) &= 50\% = \frac12, \\ P(E_2) &= 30\% = \frac{3}{10}, \\ P(E_3) &= 20\% = \frac15, \end{aligned}$$

   and the defect rates, converted from percentages, are

   $$\begin{aligned} P(D|E_1) &= 1\% = \frac{1}{100}, \\ P(D|E_2) &= 5\% = \frac{5}{100}, \\ P(D|E_3) &= 7\% = \frac{7}{100}. \end{aligned}$$
   The question does not say how fast each operator works; the intended reading (and the one behind the book's answer $\frac{5}{34}$⁠) is that items are produced at the same rate whoever is on the job, so time shares are item shares. The defect rates are measured inside each operator's own output, so they are conditional probabilities.
   $a\% = \frac{a}{100}$⁠; a rate inside one group is a conditional probability. Lesson: conditional probability

3. The chance of a defective item

   Since $E_1$⁠, $E_2$⁠, $E_3$ form a partition, by the theorem of total probability

   $$\begin{aligned} P(D) &= P(E_1)P(D|E_1) \\ &\quad + P(E_2)P(D|E_2) \\ &\quad + P(E_3)P(D|E_3) \\ &= \frac12\cdot\frac{1}{100} + \frac{3}{10}\cdot\frac{5}{100} \\ &\quad + \frac15\cdot\frac{7}{100} \\ &= \frac{1}{200} + \frac{15}{1000} + \frac{7}{500} \\ &= \frac{5}{1000} + \frac{15}{1000} + \frac{14}{1000} \\ &= \frac{34}{1000}. \end{aligned}$$
   A defective item comes from exactly one operator, so the three path probabilities add. Keeping $\frac{34}{1000}$ unsimplified makes the next step easy, because the 1000 cancels.
   Multiply along a path, add the paths. Basics: tree diagrams

4. Produced by A, given defective: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_1|D) &= \frac{P(E_1)P(D|E_1)}{P(D)} \\ &= \frac{5}{1000} \div \frac{34}{1000} \\ &= \frac{5}{1000} \times \frac{1000}{34} = \frac{5}{34}. \end{aligned}$$
   The top is operator A's path to a defective item; the bottom is all three paths. Out of every 34 parts of defect probability, only 5 belong to A. The fraction $\frac{5}{34}$ is in lowest terms, since $34 = 2 \times 17$ has no factor 5.
   Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction

5. A quick sense check

   For the other operators, $P(E_2|D) = \frac{15}{34}$ and $P(E_3|D) = \frac{14}{34} = \frac{7}{17}$⁠. Then $\frac{5}{34} + \frac{15}{34} + \frac{14}{34} = 1$⁠.

   Operator A produces half of all items but only $\frac{5}{34} \approx 0.147$ of the defective ones, because A's defect rate is the lowest.

   A defective item was produced by exactly one operator, so the three updated probabilities must add to 1.
   The posterior probabilities of a partition add to 1. Lesson: Bayes' theorem

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