An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

Total number of balls in the urn = $$3 + 4 + 2 = 9$$.

The experiment: draw 3 balls without replacement. Total equally likely ways = number of 3-element subsets of 9 balls: $$\binom{9}{3} = 84$$.

For the three balls to have different colours we need 1 red, 1 blue, and 1 green.

• Ways to choose 1 red out of 3 red balls: $$\binom{3}{1}=3$$.
• Ways to choose 1 blue out of 4 blue balls: $$\binom{4}{1}=4$$.
• Ways to choose 1 green out of 2 green balls: $$\binom{2}{1}=2$$.

By the rule of product, favourable ways $$= 3 \times 4 \times 2 = 24$$.

Therefore, required probability $$P = \frac{\text{favourable ways}}{\text{total ways}} = \frac{24}{84} = \frac{2}{7}.$$

Option A which is: $$\frac{2}{7}$$