An urn contains 9 balls of which 3 are red, 4 are blue and 2 are green. Three balls are drawn at random from…

An urn contains 9 balls of which 3 are red, 4 are blue and 2 are green. Three balls are drawn at random from the urn. The probability that the three balls have difference colours is
  1. $\frac{1}{14}$
  2. $\frac{3}{14}$
  3. $\frac{1}{7}$
  4. $\frac{2}{7}$

Solution

$\begin{aligned} \text { Required probability } & =\frac{{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^9 \mathrm{C}_3} \\ & =\frac{3 \times 4 \times 2}{\left(\frac{9 !}{3 ! 6 !}\right)}=\frac{24 \times 6}{9 \times 8 \times 7}=\frac{2}{7}\end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

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