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.

Answer 1

Let A, E_1, 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.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{100}{600} \\ P\left(E_2\right)=\frac{200}{600} \\ P\left(E_3\right)=\frac{300}{600} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{2}{100} \\ P\left(A/E_2\right)=\frac{3}{100} \\ P\left(A/E_3\right)=\frac{5}{100} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)+P\left(E_3\right)P\left(A/E_3\right)} \\ =\frac{{\displaystyle \frac{100}{600}}\times {\displaystyle \frac{2}{100}}}{{\displaystyle \frac{100}{600}}\times {\displaystyle \frac{2}{100}}+{\displaystyle \frac{200}{600}}\times {\displaystyle \frac{3}{100}+\frac{300}{600}\times \frac{5}{100}}} \\ =\frac{2}{2+{\displaystyle 6+15}}=\frac{2}{23} \end{gathered}$$