An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at…

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. $\frac{1}{3}$
  2. $\frac{2}{7}$
  3. $\frac{1}{21}$
  4. $\frac{2}{23}$

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 2024 (09 May Shift 1)

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