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.

Answer 1

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)

\[\Rightarrow P\left( A \right) = \frac{C\left( 2, 1 \right)C\left( 3, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 2, 1 \right)C\left( 5, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 2, 1 \right)C\left( 4, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 3, 1 \right)C\left( 5, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 3, 1 \right)C\left( 4, 1 \right)}{C\left( 14, 2 \right)} + \frac{C\left( 5, 1 \right)C\left( 4, 1 \right)}{C\left( 14, 2 \right)}\]

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

\[= \frac{6 + 10 + 8 + 15 + 12 + 20}{91} = \frac{71}{91} = 0 . 78\]