Question

In a factory, machine A produces 30% of the total output, machine B produces 25% and the machine C produces the remaining output. If defective items produced by machines A, B and C are 1%, 1.2%, 2% respectively. Three machines working together produce 10000 items in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine B?

Answer 1

Let A, E₁, 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.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{30}{100} \\ P\left(E_2\right)=\frac{25}{100} \\ P\left(E_3\right)=\frac{45}{100} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{1}{100} \\ P\left(A/E_2\right)=\frac{1.2}{100} \\ P\left(A/E_3\right)=\frac{2}{100} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)+P\left(E_3\right)P\left(A/E_3\right)} \\ =\frac{{\displaystyle \frac{30}{100}\times \frac{1}{100}}}{{\displaystyle \frac{30}{100}\times +\times \frac{1}{100}\frac{25}{100}\frac{1.2}{100}+\frac{45}{100}\times \frac{2}{100}}} \\ =\frac{30}{30+30+90}=\frac{30}{150}=0.2 \end{gathered}$$