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

# 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

1. A

$\frac{1}{21}$

2. B

$\frac{2}{23}$

3. C

$\frac{1}{3}$

4. D

$\frac{2}{7}$

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### Solution:

Total number of ways of drawing three balls out of $9$ is

n(S)=84

E= event of drawing 3 balls of different colours

The number of ways of drawing the balls so that balls are of different colours is

n(E)=24

probability of the required event $P\left(E\right)=\frac{24}{84}=\frac{2}{7}$

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