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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.
Solution
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
\[\therefore P\left( E_1 \right) = \frac{5}{11} \]
\[ P\left( E_2 \right) = \frac{6}{11}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{3}{8}\]
\[P\left( A/ E_2 \right) = \frac{4}{8}\]
\[\text{ Using the law of total probability, we get} \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{5}{11} \times \frac{3}{8} + \frac{6}{11} \times \frac{4}{8}\]
\[ = \frac{15}{88} + \frac{24}{88}\]
\[ = \frac{39}{88}\]
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