Question

A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.

Answer 1

Let A, E₁ and E₂ denote the events that the ball is white, bag I is chosen and bag II is chosen, respectively.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{1}{2} \\ P\left(E_2\right)=\frac{1}{2} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{1}{7} \\ P\left(A/E_2\right)=\frac{4}{7} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{7}}}{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{7}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{4}{7}}} \\ =\frac{1}{1+{\displaystyle 4}}=\frac{1}{5} \end{gathered}$$