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Question
A factory has three machines A, B and C, which produce 100, 200 and 300 items of a particular type daily. The machines produce 2%, 3% and 5% defective items respectively. One day when the production was over, an item was picked up randomly and it was found to be defective. Find the probability that it was produced by machine A.
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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.
\[\therefore P\left( E_1 \right) = \frac{100}{600}\]
\[ P\left( E_2 \right) = \frac{200}{600} \]
\[ P\left( E_3 \right) = \frac{300}{600}\]
\[\text{ Now }, \]
\[P\left( A/ E_1 \right) = \frac{2}{100}\]
\[P\left( A/ E_2 \right) = \frac{3}{100}\]
\[P\left( A/ E_3 \right) = \frac{5}{100}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{100}{600} \times \frac{2}{100}}{\frac{100}{600} \times \frac{2}{100} + \frac{200}{600} \times \frac{3}{100} + \frac{300}{600} \times \frac{5}{100}}\]
\[ = \frac{2}{2 + 6 + 15} = \frac{2}{23}\]
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