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Question
The contents of three bags I, II and III are as follows:
Bag I : 1 white, 2 black and 3 red balls,
Bag II : 2 white, 1 black and 1 red ball;
Bag III : 4 white, 5 black and 3 red balls.
A bag is chosen at random and two balls are drawn. What is the probability that the balls are white and red?
Solution
A white ball and a red ball can be drawn in three mutually exclusive ways:
(I) Selecting bag I and then drawing a white and a red ball from it
(II) Selecting bag II and then drawing a white and a red ball from it
(II) Selecting bag III and then drawing a white and a red ball from it
Let E1, E2 and A be the events as defined below:
E1 = Selecting bag I
E2 = Selecting bag II
E3 = Selecting bag II
A = Drawing a white and a red ball
It is given that one of the bags is selected randomly.
\[\therefore P\left( E_1 \right) = \frac{1}{3}\]
\[ P\left( E_2 \right) = \frac{1}{3}\]
\[ P\left( E_3 \right) = \frac{1}{3}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{{}^1 C_1 \times^3 C_1}{{}^6 C_2} = \frac{3}{15}\]
\[P\left( A/ E_2 \right) = \frac{{}^2 C_1 \times^1 C_1}{{}^4 C_2} = \frac{2}{6}\]
\[P\left( A/ E_3 \right) = \frac{{}^4 C_1 \times^3 C_1}{{}^{12} C_2} = \frac{12}{66}\]
\[\text{ Using the law of total probability, we get} \]
\[\text{ Required probability} = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)\]
\[ = \frac{1}{3} \times \frac{3}{15} + \frac{1}{3} \times \frac{2}{6} + \frac{1}{3} \times \frac{12}{66}\]
\[ = \frac{1}{15} + \frac{1}{9} + \frac{2}{33}\]
\[ = \frac{33 + 55 + 30}{495} = \frac{118}{495}\]
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