In a bolt factory, machines A, B and C manufacture 60%, 25%and 15% respectively. Of the total of their output 1%, 2%, 1% is defective bolts. A bolt is drawn at random from the total production and found to be defective. From which machine, the defective bolt is most expected to have been manufactured   

Let $E_1,E_2,E_3$ be the events of drawing a bolt produced by machine A, B and C respectively. Let D the event of drawing a defective bolt. We have $P\left(E_1\right)=\frac{60}{100},P\left(E_2\right)=\frac{25}{100},P\left(E_3\right)=\frac{15}{100}$

$P\left(D/E_1\right)=\frac{1}{100},P\left(D/E_2\right)=\frac{2}{100},P\left(D/E_3\right)=\frac{1}{100}$

 The events $E_1,E_2$ and $E_3$ are mutually exclusive and exhaustive.

P(defective bolt is produced by A)

$=P\left(E_1/D\right)=\frac{P\left(E_1\right)P\left(D/E_1\right)}{P\left(E_1\right)P\left(D/E_1\right)+P\left(E_2\right)P\left(D/E_2\right)+P\left(E_3\right)P\left(D/E_3\right)}$

                          $=\frac{\frac{60}{100}\times \frac{1}{100}}{\frac{60}{100}\times \frac{1}{100}+\frac{25}{100}\times \frac{2}{100}+\frac{15}{100}\times \frac{1}{100}}$

$=\frac{60}{125}=\frac{12}{25}$ [using baye’s theorem]

Similarly, P(defective bolt is produced by B)

$=P\left(E_2/D\right)=\frac{P\left(E_2\right)P\left(D/E_2\right)}{P\left(E_1\right)P\left(D/E_1\right)+P\left(E_2\right)P\left(D/E_2\right)+P\left(E_3\right)P\left(D/E_3\right)}$

$=\frac{\frac{25}{100}\times \frac{2}{100}}{\frac{60}{100}\times \frac{1}{100}+\frac{25}{100}\times \frac{2}{100}+\frac{15}{100}\times \frac{1}{100}}=\frac{50}{125}=\frac{10}{25}$

P(defective bolt is produced by C)

$=P\left(E_3/D\right)=\frac{P\left(E_3\right)P\left(D/E_3\right)}{P\left(E_1\right)P\left(D/E_1\right)+P\left(E_2\right)P\left(D/E_2\right)+P\left(E_3\right)P\left(D/E_3\right)}$

                        $=\frac{\frac{15}{100}\times \frac{1}{100}}{\frac{60}{100}\times \frac{1}{100}+\frac{25}{100}\times \frac{2}{100}+\frac{15}{100}\times \frac{1}{100}}=\frac{15}{125}=\frac{3}{25}$

From the above 3 probabilities it is clear that the machine A has the highest probability of producing a defective bolt.