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.

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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)

$\Rightarrow P (A) = \frac{C (2, 1) C (3, 1)}{C (14, 2)} + \frac{C (2, 1) C (5, 1)}{C (14, 2)} + \frac{C (2, 1) C (4, 1)}{C (14, 2)} + \frac{C (3, 1) C (5, 1)}{C (14, 2)} + \frac{C (3, 1) C (4, 1)}{C (14, 2)} + \frac{C (5, 1) C (4, 1)}{C (14, 2)}$

$= \frac{2 \times 3}{91} + \frac{2 \times 5}{91} + \frac{2 \times 4}{91} + \frac{3 \times 5}{91} + \frac{3 \times 4}{91} + \frac{5 \times 4}{91}$

$= \frac{6 + 10 + 8 + 15 + 12 + 20}{91} = \frac{71}{91} = 0.78$

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