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प्रश्न
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?
उत्तर
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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