Example 24: Machine set-up: correct, given 2 acceptable items, $P = \frac{81}{85}$

Basics you need

Solution

1. Name the events

   Let

   • $A$⁠: the machine produces 2 acceptable items;
   • $B_1$⁠: the set-up is correct;
   • $B_2$⁠: the set-up is incorrect.

   $B_2$ is the complement of $B_1$⁠, so $B_1$ and $B_2$ form a partition of the sample space.

   The set-up is the hidden cause; the two acceptable items are what we observe. We want the probability of the cause "correct set-up" after seeing them.
   $\{B_1, B_2\}$ is a partition. Lesson: Bayes' theorem

2. The probabilities of the causes

   $$\begin{aligned} P(B_1) &= 80\% = \frac45, \\ P(B_2) &= 1 - \frac45 = \frac15. \end{aligned}$$
   "80% of the set ups are correctly done" is the chance of a correct set-up before we see any items. The book writes these as 0.8 and 0.2.
   $P(B_2) = 1 - P(B_1)$⁠. Basics: rules of probability

3. The chance of 2 acceptable items for each set-up

   After a correct set-up each item is acceptable with probability $\frac{9}{10}$⁠; after an incorrect one with probability $\frac{4}{10}$⁠. Taking the two items as independent of each other (given the set-up),

   $$\begin{aligned} P(A|B_1) &= \frac{9}{10} \times \frac{9}{10} = \frac{81}{100}, \\ P(A|B_2) &= \frac{4}{10} \times \frac{4}{10} = \frac{16}{100}. \end{aligned}$$
   The question gives the rate for one item; we need both items acceptable. The book multiplies the rates ($0.9 \times 0.9$ and $0.4 \times 0.4$⁠), which assumes that, once the set-up is fixed, whether one item is acceptable does not affect the other. The question does not say this in words, but it is the natural reading and the book's.
   Independent: $P(\text{both}) = P(\text{first})\,P(\text{second})$⁠. Lesson: independent events

4. The chance of 2 acceptable items: total probability

   By the theorem of total probability,

   $$\begin{aligned} P(A) &= P(B_1)P(A|B_1) \\ &\quad + P(B_2)P(A|B_2) \\ &= \frac45\cdot\frac{81}{100} + \frac15\cdot\frac{16}{100} \\ &= \frac{324}{500} + \frac{16}{500} = \frac{340}{500}. \end{aligned}$$
   The two paths cannot both happen, so they add. In decimals they are the book's $0.8 \times 0.9 \times 0.9 = 0.648$ and $0.2 \times 0.4 \times 0.4 = 0.032$⁠, with sum 0.68.
   Multiply along a path, add the paths. Basics: tree diagrams

5. Correct set-up, given 2 acceptable items: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(B_1|A) &= \frac{P(B_1)P(A|B_1)}{P(A)} \\ &= \frac{324}{500} \div \frac{340}{500} = \frac{324}{340} \\ &= \frac{81}{85} \approx 0.9529. \end{aligned}$$
   The top is the correct-set-up path; the bottom is all ways of getting 2 acceptable items. The 500 cancels, and $324 = 4 \times 81$⁠, $340 = 4 \times 85$⁠. $81 = 3^4$ and $85 = 5 \times 17$ share no factor. The book writes $\frac{648}{680} = 0.95$⁠: $\frac{648}{680}$ is the same fraction ($648 = 8 \times 81$⁠, $680 = 8 \times 85$⁠); its exact value is $\frac{81}{85} \approx 0.9529$⁠, and 0.95 is that value rounded to two decimal places.
   Simplify by the highest common factor. Basics: working with fractions

6. What the answer means

   Before any items were seen, a correct set-up had probability $\frac45 = 0.8$⁠. Two acceptable items raise it to $\frac{81}{85} \approx 0.95$⁠. An incorrect set-up keeps $P(B_2|A) = \frac{16}{340} = \frac{4}{85}$⁠, and $\frac{81}{85} + \frac{4}{85} = 1$⁠.

   Two acceptable items are about 5 times as likely after a correct set-up as after an incorrect one ($\frac{81}{100}$ against $\frac{16}{100}$⁠), so they shift the odds strongly towards a correct set-up.
   For a partition, $P(B_1|A) + P(B_2|A) = 1$⁠. Lesson: Bayes' theorem

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