Question

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced. What is the probability that it was produced by A?

Answer 1

Let E₁, E₂ and E₃ be the time taken by machine operators A, B, and C, respectively.

Let X be the event of producing defective items.

\[\therefore P\left( E_1 \right) = 50 % = \frac{1}{2} \]

\[ P\left( E_2 \right) = 30 % = \frac{3}{10}\]

\[ P\left( E_3 \right) = 20 % = \frac{1}{5}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = 1 % = \frac{1}{100}\]

\[P\left( A/ E_2 \right) = 5 % = \frac{5}{100}\]

\[P\left( A/ E_2 \right) = 7 % = \frac{7}{100}\]

\[\text{ Using Bayes' theorem, we get } \]

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

\[ = \frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{2} \times \frac{1}{100} + \frac{3}{10} \times \frac{5}{100} + \frac{1}{5} \times \frac{7}{100}}\]

\[ = \frac{5}{34}\]