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 colour is

A urn has 3 red balls, 4 blue balls and 2 green balls.
Event of choosing 3 balls
n(S) = 9C3
Event of choosing 3 different colour balls
n(E) = 3C1 × 4C1 × 2C1
Probability = (3 × 4 × 2) / 9C3 = 24 × 3! / (9! × 6!) = 24 × 6 / (9 × 8 × 7) = 24 × 6 / 504 = 144 / 504 = 2/7
9C3 = 9!/(6!3!) = (9×8×7)/(3×2×1) = 84
Probability = 24/84 = 2/7
However, if the question is asking for the probability expressed as a fraction, the solution is 2/7. If the options are interpreted as fractions, then none of the given options is correct. If the options are interpreted as decimals rounded to a certain number of places, the closest answer is 27, but the solution is not 27. The correct calculation is:
Probability = (3C1 * 4C1 * 2C1) / 9C3 = (3 * 4 * 2) / (9! / (3! * 6!)) = 24 / 84 = 2/7 ≈ 0.2857
There seems to be a discrepancy between the given solution and the expected solution. The provided solution appears to have calculation errors.