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?

 

Answer 1

Let A, E₁ and E₂ denote the events that the item is defective, machine A is selected and machine B is selected, respectively.

\[\therefore P\left( E_1 \right) = \frac{60}{100} \]

\[ P\left( E_2 \right) = \frac{40}{100}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{2}{100}\]

\[P\left( A/ E_2 \right) = \frac{1}{100}\]

\[\text{ Using the law of total probability, we get} \]

\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]

\[ = \frac{60}{100} \times \frac{2}{100} + \frac{40}{100} \times \frac{1}{100}\]

\[ = \frac{120}{10000} + \frac{40}{10000}\]

\[ = \frac{120 + 40}{10000} = \frac{160}{10000} = 0 . 016\]