Question

A factory has three machines X, Y and Z producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% and Zproduces 2% defective bolts. At the end of a day, a bolt is drawn at random and is found to be defective. What is the probability that this defective bolt has been produced by machine X?

 

Answer 1

Let E₁, E₂ and E₃ denote the events that machine X produces bolts, machine Yproduces bolts and machine Z produces bolts, respectively.

Let A be the event that the bolt is defective.

Total number of bolts = 1000 + 2000 + 3000 = 6000

P(E₁) = \[\frac{1000}{6000} = \frac{1}{6}\]

P(E₂) =\[\frac{2000}{6000} = \frac{1}{3}\]=

P(E₃) =\[\frac{3000}{6000} = \frac{1}{2}\]

The probability that the defective bolt is produced by machine X is given by P (E₁/A).

\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = 1 \% = \frac{1}{100}\]
\[P\left( A/ E_2 \right) = 1 . 5 \% = \frac{15}{1000}\]
\[P\left( A/ E_3 \right) = 2 \% = \frac{2}{100}\]
\[\text{ Using Bayes' theorem, we get } \]
\[ \text{ Required probability }  = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100} + \frac{1}{3} \times \frac{15}{1000} + \frac{1}{2} \times \frac{2}{100}}\]
\[ = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{2} + 1} = \frac{\frac{1}{6}}{\frac{1 + 3 + 6}{6}} = \frac{1}{10}\]