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Question
A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is
Options
2/15
7/15
8/15
14/15
Solution
8/15
A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag X and then drawing a white ball from it.
(II) Selecting bag Y and then drawing a white ball from it.
Let E1, E2 and A be three events as defined below:
E1 = Selecting bag X
E2 = Selecting bag Y
A = Drawing a white ball
We know that one bag is selected randomly.
\[\therefore P\left( E_1 \right) = \frac{1}{2} \]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[ P\left( A/ E_1 \right) = \frac{2}{5}\]
\[P\left( A/ E_2 \right) = \frac{4}{6} = \frac{2}{3}\]
\[\text{ Using the law of total probability, we get} \]
\[\text{ Required probability } = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]
\[ = \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{2}{3}\]
\[ = \frac{1}{5} + \frac{1}{3}\]
\[ = \frac{3 + 5}{15} = \frac{8}{15}\]
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