Question

One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.

Answer 1

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

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

Let E₁, E₂ and A be the events as defined below:

E₁ = A red ball is transferred from first to second bag
E₂ = A yellow ball is transferred from first to second bag
A =  A yellow ball is drawn

\[\therefore P\left( E_1 \right) = \frac{5}{9} \]

\[ P\left( E_2 \right) = \frac{4}{9}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{6}{10}\]

\[P\left( A/ E_2 \right) = \frac{7}{10}\]

\[\text{ Using the law of total probability, we get } \]

\[\text{ Required probability }  = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]

\[ = \frac{5}{9} \times \frac{6}{10} + \frac{4}{9} \times \frac{7}{10}\]

\[ = \frac{30}{90} + \frac{28}{90}\]

\[ = \frac{58}{90} = \frac{29}{45}\]