Editing Bayes' theorem (section)

===Defective item rate===
{| class="wikitable floatright"
! {{diagonal split header|<br />Machine|Condition}}
! style="width:5ex;"|Defective
! style="width:5ex;"|Flawless
! rowspan="6" |
! Total
|-
! A
| style="text-align:right" | 10
| style="text-align:right" | 190
| style="text-align:right" | 200
|-
! B
| style="text-align:right" | 9
| style="text-align:right" | 291
| style="text-align:right" | 300
|-
! C
| style="text-align:right" | '''5'''
| style="text-align:right" | 495
| style="text-align:right" | 500
|-
| colspan="5" style="padding:0;"|
|-
! Total
| style="text-align:right" | '''24'''
| style="text-align:right" | 976
| style="text-align:right" | 1000
|}

A factory produces items using three machines—A, B, and C—which account for 20%, 30%, and 50% of its output, respectively. Of the items produced by machine A, 5% are defective, while 3% of B's items and 1% of C's are defective. If a randomly selected item is defective, what is the probability it was produced by machine C?

Once again, the answer can be reached without using the formula by applying the conditions to a hypothetical number of cases. For example, if the factory produces 1,000 items, 200 will be produced by A, 300 by B, and 500 by C. Machine A will produce 5% &times; 200 = 10 defective items, B 3% &times; 300 = 9, and C 1% &times; 500 = 5, for a total of 24. Thus 24/1000 (2.4%) of the total output will be defective and the likelihood that a randomly selected defective item was produced by machine C is 5/24 (~20.83%).

This problem can also be solved using Bayes' theorem: Let ''X<sub>i</sub>'' denote the event that a randomly chosen item was made by the ''i'' <sup>th</sup> machine (for ''i''&nbsp;= A,B,C). Let ''Y'' denote the event that a randomly chosen item is defective. Then, we are given the following information:
:<math>P(X_A) = 0.2, \quad P(X_B) = 0.3, \quad  P(X_C) = 0.5.</math>
If the item was made by the first machine, then the probability that it is defective is 0.05; that is, ''P''(''Y''&thinsp;|&thinsp;''X''<sub>A</sub>) = 0.05. Overall, we have

:<math>P(Y| X_A) = 0.05, \quad  P(Y |X_B) = 0.03, \quad  P(Y| X_C) = 0.01.</math>

To answer the original question, we first find ''P''(Y). That can be done in the following way:

:<math>P(Y) = \sum_i  P(Y| X_i) P(X_i) = (0.05)(0.2) + (0.03)(0.3) + (0.01)(0.5) = 0.024.</math>

Hence, 2.4% of the total output is defective.

We are given that ''Y'' has occurred and we want to calculate the conditional probability of ''X''<sub>C</sub>. By Bayes' theorem,

:<math>P(X_C|Y) = \frac{P(Y | X_C) P(X_C)}{P(Y)} = \frac{0.01 \cdot 0.50}{0.024} = \frac{5}{24}</math>

Given that the item is defective, the probability that it was made by machine C is 5/24. C produces half of the total output but a much smaller fraction of the defective items. Hence the knowledge that the item selected was defective enables us to replace the prior probability ''P''(''X''<sub>C</sub>)&nbsp;= 1/2 by the smaller posterior probability ''P''(X<sub>C</sub>&thinsp;|&thinsp;''Y'')&nbsp;= 5/24.