Question

A bag contains 4 white and 5 black balls and another bag contains 3 white and 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.

Answer 1

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a white ball
(II) By transferring a white ball from first to second bag, then drawing a white ball

Let E₁, E₂ and A be the events as defined below:
E₁ = A black ball is transferred from first to second bag
E₂ = A white ball is transferred from first to second bag
A = A white ball is drawn

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{5}{9} \\ P\left(E_2\right)=\frac{4}{9} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{3}{8} \\ P\left(A/E_2\right)=\frac{4}{8}=\frac{1}{2} \\ \mathrm{Using}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{total}\mathrm{probability},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right) \\ =\frac{5}{9}\times \frac{3}{8}+\frac{4}{9}\times \frac{1}{2} \\ =\frac{15}{72}+\frac{2}{9} \\ =\frac{15+16}{72}=\frac{31}{72} \end{gathered}$$