Question

A factory has two machines A and B. Past records show that the machine A produced 60% of the items of output and machine B produced 40% of the items. Further 2% of the items produced by machine A were defective and 1% produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?

Answer 1

Let A, E₁ and E₂ denote the events that the item is defective, machine A is selected and machine B is selected, respectively.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{60}{100} \\ P\left(E_2\right)=\frac{40}{100} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{2}{100} \\ P\left(A/E_2\right)=\frac{1}{100} \\ \mathrm{Using}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{total}\mathrm{probability},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right) \\ =\frac{60}{100}\times \frac{2}{100}+\frac{40}{100}\times \frac{1}{100} \\ =\frac{120}{10000}+\frac{40}{10000} \\ =\frac{120+40}{10000}=\frac{160}{10000}=0.016 \end{gathered}$$