Question

A bag A contains 5 white and 6 black balls. Another bag B contains 4 white and 3 black balls. A ball is transferred from bag A to the bag B and then a ball is taken out of the second bag. Find the probability of this ball being black.

Answer 1

A black ball can be drawn in two mutually exclusive ways:
(I) By transferring a white ball from bag A to bag B, then drawing a black ball
(II) By transferring a black ball from bag A to bag B, then drawing a black ball
Let E₁, E₂ and A be the events as defined below:
E₁ = A white ball is transferred from bag A to bag B
E₂ = A black ball is transferred from bag A to bag B
A = A black ball is drawn

\[\therefore P\left( E_1 \right) = \frac{5}{11} \]

\[ P\left( E_2 \right) = \frac{6}{11}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{3}{8}\]

\[P\left( A/ E_2 \right) = \frac{4}{8}\]

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

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

\[ = \frac{5}{11} \times \frac{3}{8} + \frac{6}{11} \times \frac{4}{8}\]

\[ = \frac{15}{88} + \frac{24}{88}\]

\[ = \frac{39}{88}\]