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Question
A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.
Solution
A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag I and then drawing a white ball from it
(II) Selecting bag II and then drawing a white ball from it
Let E1, E2 and A be the events as defined below:
E1 = Selecting bag I
E2 = Selecting bag II
A = Drawing a white ball
It is given that one of the bags is selected randomly.
\[\therefore P\left( E_1 \right) = \frac{1}{2} \]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[ \text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{3}{5}\]
\[P\left( A/ E_2 \right) = \frac{2}{6} = \frac{1}{3}\]
\[\text{ Using the law of total probability, we get } \]
\[\text{ Required probability} = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{3}\]
\[ = \frac{3}{10} + \frac{1}{6}\]
\[ = \frac{9 + 5}{30} = \frac{14}{30} = \frac{7}{15}\]
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