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Question
Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colours.
Solution
Out of 14 balls, two balls can be chosen in 14C2 ways.
So, favourable number of elementary events = 14C2
Let A be the event of getting two balls of different colours.
∴ P(A) = P(1 white and 1 red) + P(1 white and 1 green) + P(1 white and 1 black) + P(1 red and 1 green)
+ P(1 red and 1 black) + P(1 green and 1 black)
\[\Rightarrow P\left( A \right) = \frac{C\left( 2, 1 \right)C\left( 3, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 2, 1 \right)C\left( 5, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 2, 1 \right)C\left( 4, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 3, 1 \right)C\left( 5, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 3, 1 \right)C\left( 4, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 5, 1 \right)C\left( 4, 1 \right)}{C\left( 14, 2 \right)}\]
\[= \frac{2 \times 3}{91} + \frac{2 \times 5}{91} + \frac{2 \times 4}{91} + \frac{3 \times 5}{91} + \frac{3 \times 4}{91} + \frac{5 \times 4}{91}\]
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