Question

A bag contains 6 red and 8 black balls and another bag contains 8 red and 6 black balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag. Find the probability that the ball drawn is red in colour.

Answer 1

A red ball can be drawn in two mutually exclusive ways:
(I) By transferring a black ball from first to second bag, then drawing a red ball
(II) By transferring a red ball from first to second bag, then drawing a red 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 red ball is transferred from first to second bag
A = A red ball is drawn

\[\therefore P\left( E_1 \right) = \frac{8}{14} \]
\[ P\left( E_2 \right) = \frac{6}{14}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{8}{15}\]
\[P\left( A/ E_2 \right) = \frac{9}{15}\]
\[\text{ Using the law of total probability, we get } \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{8}{14} \times \frac{8}{15} + \frac{6}{14} \times \frac{9}{15}\]
\[ = \frac{64}{210} + \frac{54}{210}\]
\[ = \frac{118}{210} = \frac{59}{105}\]