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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.
Solution
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
\[\therefore P\left( E_1 \right) = \frac{5}{9} \]
\[ P\left( E_2 \right) = \frac{4}{9}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{6}{10}\]
\[P\left( A/ E_2 \right) = \frac{7}{10}\]
\[\text{ Using the law of total probability, we get } \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{5}{9} \times \frac{6}{10} + \frac{4}{9} \times \frac{7}{10}\]
\[ = \frac{30}{90} + \frac{28}{90}\]
\[ = \frac{58}{90} = \frac{29}{45}\]
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