Chapter 13 · Probability

Example 19: Defective bolt: made by machine B, $P = \frac{28}{69}$

ExamplesBayes' theorem · Section 13.5

The question

In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

Exam answer

Let $B_1$ ⁠, $B_2$ ⁠, $B_3$ ⁠: the bolt is manufactured by machine A, B, C, respectively, and $E$ ⁠: the bolt is defective.

$B_1$ ⁠, $B_2$ and $B_3$ form a partition of the sample space (each bolt is made by exactly one machine).

$P(B_1) = 25\% = \frac{25}{100}$

$P(B_2) = 35\% = \frac{35}{100}$

$P(B_3) = 40\% = \frac{40}{100}$

$P(E|B_1) = 5\% = \frac{5}{100}$

$P(E|B_2) = 4\% = \frac{4}{100}$

$P(E|B_3) = 2\% = \frac{2}{100}$

By the theorem of total probability,

\begin{aligned} P(E) &= P(B_1)P(E|B_1) \\ &\quad + P(B_2)P(E|B_2) \\ &\quad + P(B_3)P(E|B_3) \\ &= \frac{25}{100}\cdot\frac{5}{100} \\ &\quad + \frac{35}{100}\cdot\frac{4}{100} \\ &\quad + \frac{40}{100}\cdot\frac{2}{100} \\ &= \frac{125 + 140 + 80}{10000} \\ &= \frac{345}{10000} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(B_2|E) &= \frac{P(B_2)P(E|B_2)}{P(E)} \\ &= \frac{140}{10000} \div \frac{345}{10000} \\ &= \frac{140}{345} \end{aligned}

$\therefore$ The probability that the defective bolt is manufactured by machine B is $\dfrac{28}{69}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Weigh each machine by how much it makes and how often it fails. A defective bolt comes from A, B or C. Machine B's share of the defective bolts is

\frac{P(\text{made by B and defective})}{P(\text{defective})},

and each "made by ... and defective" is a product: the machine's share of output times its defect rate. That is Bayes' theorem.

Given, to find, and the plan

Given: Shares of output: A 25%, B 35%, C 40%. Defective: 5% of A's bolts, 4% of B's, 2% of C's. The bolt drawn is defective.
To find: The probability that it was made by machine B: $P(B_2|E)$ ⁠.
Plan: (1) Name the events; check the partition. (2) Percentages as fractions. (3) Total probability for $P(E)$ ⁠. (4) Bayes' theorem; simplify.

Basics you need

1Percentages as fractions

$x\% = \frac{x}{100}$ ⁠: $35\% = \frac{35}{100}$ ⁠. Basics: working with fractions

2Defect rates are conditional probabilities (Section 13.2)

"Of their outputs, 4 percent (of B's) are defective" is $P(\text{defective} \mid \text{made by B})$ ⁠. Lesson: conditional probability

3Theorem of total probability (Section 13.5.2)

$P(E) = \sum P(B_i)P(E|B_i)$ over a partition. Lesson: total probability and Bayes' theorem

4Bayes' theorem (Section 13.5)

$P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(E)}$ ⁠. Lesson: Bayes' theorem

Solution

Name the events

Let
- $B_1$ ⁠: the bolt is manufactured by machine A;
- $B_2$ ⁠: the bolt is manufactured by machine B;
- $B_3$ ⁠: the bolt is manufactured by machine C;
- $E$ ⁠: the bolt is defective.
$B_1$ ⁠, $B_2$ ⁠, $B_3$ are mutually exclusive and exhaustive (each bolt is made by exactly one machine), and each has nonzero probability, so they form a partition of the sample space.

The machines are the causes; the defect is the effect we see. The book names these events $B_1, B_2, B_3$ ⁠, and we keep its names. Take care not to mix up the machine called B with the event names: the event "made by machine B" is $B_2$ ⁠.

A partition: pairwise disjoint, exhaustive, nonzero probabilities. Lesson: Bayes' theorem
Every number as a fraction
$\begin{aligned} P(B_1) &= 25\% = \frac{25}{100}, \\ P(B_2) &= 35\% = \frac{35}{100}, \\ P(B_3) &= 40\% = \frac{40}{100}, \\ P(E|B_1) &= 5\% = \frac{5}{100}, \\ P(E|B_2) &= 4\% = \frac{4}{100}, \\ P(E|B_3) &= 2\% = \frac{2}{100}. \end{aligned}$
A bolt drawn at random from the whole product comes from machine B with probability equal to B's share, 35%. "Of their outputs, 5, 4 and 2 percent are respectively defective" gives the defect rate of each machine, so these are $P(E|B_i)$ ⁠. Keeping the denominator 100 (rather than $\frac14$ ⁠, $\frac{7}{20}$ ⁠, $\frac25$ ⁠) makes every product below have the same denominator 10000. The three shares add to $\frac{100}{100} = 1$ ⁠, as they must for a partition.

$x\% = \frac{x}{100}$ ⁠. Basics: working with fractions
The chance of a defective bolt: total probability

By the theorem of total probability,
$\begin{aligned} P(E) &= P(B_1)P(E|B_1) \\ &\quad + P(B_2)P(E|B_2) \\ &\quad + P(B_3)P(E|B_3) \\ &= \frac{25}{100}\cdot\frac{5}{100} + \frac{35}{100}\cdot\frac{4}{100} \\ &\quad + \frac{40}{100}\cdot\frac{2}{100} \\ &= \frac{125}{10000} + \frac{140}{10000} + \frac{80}{10000} \\ &= \frac{345}{10000}. \end{aligned}$

The tree for Example 19. The defective paths have products $\frac{125}{10000}$ (A), $\frac{140}{10000}$ (B, rose) and $\frac{80}{10000}$ (C); together $\frac{345}{10000}$ ⁠.

A defective bolt came from exactly one machine, so the three paths are mutually exclusive and their probabilities add. In decimals these are the book's numbers: $P(E) = 0.0345$ and the B path is $0.0140$ ⁠.

Multiply along a path, add the paths. Basics: tree diagrams
Made by B, given defective: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(B_2|E) &= \frac{P(B_2)P(E|B_2)}{P(E)} \\ &= \frac{140}{10000} \div \frac{345}{10000} \\ &= \frac{140}{345} = \frac{28}{69}. \end{aligned}$
The top is the B path, the bottom is all defective bolts; the common denominator 10000 cancels. Then divide 140 and 345 by their highest common factor 5: $140 = 5 \times 28$ ⁠, $345 = 5 \times 69$ ⁠. $28 = 2^2 \times 7$ and $69 = 3 \times 23$ share no factor. This is the book's $\frac{0.0140}{0.0345} = \frac{28}{69}$ ⁠.

Bayes' theorem: one path over all the paths. Lesson: Bayes' theorem
What the answer means

The other two machines get $P(B_1|E) = \frac{125}{345} = \frac{25}{69}$ and $P(B_3|E) = \frac{80}{345} = \frac{16}{69}$ ⁠, and $\frac{25 + 28 + 16}{69} = 1$ ⁠.

Machine C makes the most bolts (40%) but has the lowest defect rate, so it is the least likely source of a defective bolt. Machine B makes 35% of the bolts but about 40.6% of the defective ones.

Each machine's chance is proportional to (share of output) × (defect rate): $125 : 140 : 80$ ⁠. Comparing these products tells you at once which cause is most likely.

For a partition, the $P(B_i|E)$ add to 1. Lesson: Bayes' theorem

Result

The probability that the defective bolt was manufactured by machine B is $\frac{28}{69}$ ⁠.

(Textbook answer: $\frac{28}{69}$ ⁠.)

Check it with numbers

Picture 10000 bolts. Machine A makes 2500, of which 5%, 125, are defective; B makes 3500, of which 4%, 140, are defective; C makes 4000, of which 2%, 80, are defective. Of the $125 + 140 + 80 = 345$ defective bolts, 140 are from B: $\frac{140}{345} \approx 0.4058$ ⁠, the same as $\frac{28}{69} \approx 0.4058$ ⁠.

Common mistakes

Answering 0.04 or 0.35. 4% is $P(E|B_2)$ and 35% is $P(B_2)$ ⁠; the question asks for $P(B_2|E)$ ⁠, which needs both and the other machines too.
Mixing up the event $B_2$ and machine B. In the book's notation machine B is $B_2$ ⁠, not $B_1$ ⁠.
Adding the defect rates. $5\% + 4\% + 2\%$ is not $P(E)$ ⁠; each rate must be weighted by its machine's share.

What to take away

With several machines (or causes), $P(\text{cause} \mid \text{effect})$ is proportional to $P(\text{cause}) \times P(\text{effect} \mid \text{cause})$ ⁠.
Keeping a common denominator (here 10000) makes Bayes' theorem a ratio of whole numbers.
Two-machine and three-operator versions: Exercise 13.3, Question 8 and Question 11.

Every fact used, and where it comes from

Fact	Used in	Source
Mutually exclusive and exhaustive events form a partition	Step 1	Section 13.5.1 (Lesson: Bayes' theorem)
$x\% = \frac{x}{100}$ ⁠; conditional probability	Step 2	Section 13.2 (Lesson: conditional probability)
Theorem of total probability	Step 3	Section 13.5.2 (Lesson: total probability)
Bayes' theorem	Steps 4, 5	Section 13.5 (Examples 15 to 21)
Cancelling a common factor	Step 4	Arithmetic (Basics: working with fractions)

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Example 19: Defective bolt: made by machine B, P=2869P = \frac{28}{69}P=6928​

Basics you need

Solution

Name the events

Every number as a fraction

The chance of a defective bolt: total probability

Made by B, given defective: Bayes' theorem

What the answer means

Every fact used, and where it comes from

Example 19: Defective bolt: made by machine B, $P = \frac{28}{69}$