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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.

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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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Chapter 30: Probability - Exercise 31.7 [Page 97]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 30 Probability
Exercise 31.7 | Q 20 | Page 97

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