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

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