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Question
In a factory, machine A produces 30% of the total output, machine B produces 25% and the machine C produces the remaining output. If defective items produced by machines A, B and C are 1%, 1.2%, 2% respectively. Three machines working together produce 10000 items in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine B?
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Solution
Let A, E1, E2 and E3 denote the events that the item is defective, machine A is chosen, machine B is chosen and machine C is chosen, respectively.
\[\therefore P\left( E_1 \right) = \frac{30}{100}\]
\[ P\left( E_2 \right) = \frac{25}{100} \]
\[ P\left( E_3 \right) = \frac{45}{100}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{1}{100}\]
\[P\left( A/ E_2 \right) = \frac{1 . 2}{100}\]
\[P\left( A/ E_3 \right) = \frac{2}{100}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{30}{100} \times \frac{1}{100}}{\frac{30}{100} \times + \times \frac{1}{100}\frac{25}{100}\frac{1 . 2}{100} + \frac{45}{100} \times \frac{2}{100}}\]
\[ = \frac{30}{30 + 30 + 90} = \frac{30}{150} = 0 . 2\]
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