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Question
A factory has two machines A and B. Past records show that the machine A produced 60% of the items of output and machine B produced 40% of the items. Further 2% of the items produced by machine A were defective and 1% produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?
Solution
Let A, E1 and E2 denote the events that the item is defective, machine A is selected and machine B is selected, respectively.
\[\therefore P\left( E_1 \right) = \frac{60}{100} \]
\[ P\left( E_2 \right) = \frac{40}{100}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{2}{100}\]
\[P\left( A/ E_2 \right) = \frac{1}{100}\]
\[\text{ Using the law of total probability, we get} \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{60}{100} \times \frac{2}{100} + \frac{40}{100} \times \frac{1}{100}\]
\[ = \frac{120}{10000} + \frac{40}{10000}\]
\[ = \frac{120 + 40}{10000} = \frac{160}{10000} = 0 . 016\]
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