Exercise 13.3 · Q8: Two machines: defective item made by B, $P = \frac14$

Basics you need

Solution

1. Name the events

   Let

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

   $E_1$ and $E_2$ form a partition of the sample space (all the items in the stockpile): no item is made by both machines, every item is made by one of them, and each machine makes some items.

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

2. The four probabilities as fractions

   An item picked at random from the stockpile was made by A with probability 60% and by B with probability 40%:

   $$\begin{aligned} P(E_1) &= 60\% = \frac35, \\ P(E_2) &= 40\% = \frac25. \end{aligned}$$

   The defect rates of the two machines are

   $$\begin{aligned} P(D|E_1) &= 2\% = \frac{1}{50}, \\ P(D|E_2) &= 1\% = \frac{1}{100}. \end{aligned}$$
   "Chosen at random" from the stockpile means every item is equally likely, so a machine's share of the output is the probability that the item came from it. The defect rates are measured inside each machine'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 $E_1$ and $E_2$ form a partition, by the theorem of total probability

   $$\begin{aligned} P(D) &= P(E_1)P(D|E_1) \\ &\quad + P(E_2)P(D|E_2) \\ &= \frac35\cdot\frac{1}{50} + \frac25\cdot\frac{1}{100} \\ &= \frac{3}{250} + \frac{1}{250} = \frac{4}{250}. \end{aligned}$$
   A defective item comes from exactly one machine, so the two path probabilities add. Here $\frac25\cdot\frac{1}{100} = \frac{2}{500} = \frac{1}{250}$⁠. Keeping $\frac{4}{250}$ unsimplified (it equals $\frac{2}{125}$⁠) makes the next step easy, because the 250 cancels.
   Multiply along a path, add the paths. Basics: tree diagrams

4. Made by B, given defective: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_2|D) &= \frac{P(E_2)P(D|E_2)}{P(D)} \\ &= \frac{1}{250} \div \frac{4}{250} \\ &= \frac{1}{250} \times \frac{250}{4} = \frac14. \end{aligned}$$
   The top is machine B's path to a defective item; the bottom is both paths. Out of every 4 parts of defect probability, 1 belongs to machine B and 3 to machine A.
   Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction

5. A quick sense check

   For machine A, $P(E_1|D) = \frac{3}{250} \div \frac{4}{250} = \frac34$⁠, and $\frac14 + \frac34 = 1$⁠.

   Machine B makes $\frac25$ of all items but only $\frac14$ of the defective ones, because its defect rate is half of A's.

   A defective item was made by exactly one machine, so the two updated probabilities must add to 1. Learning that the item is defective lowers B's chance from $\frac25 = 0.4$ to $\frac14 = 0.25$⁠.
   The posterior probabilities of a partition add to 1. Lesson: Bayes' theorem

Every fact used, and where it comes from