Question

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. $\frac{2}{7}$
• B. $\frac{1}{21}$
• C. $\frac{2}{23}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

$\frac{2}{7}$

A urn has $3$ are red,$4$ are blue and $2$ are green

event of choosing 3 balls

$n\left(S\right){=}^9{C}_3$

event of choosing 3 different colour balls

$n\left(E\right){=}^3{C}_1{\times }^4{C}_1{\times }^2{C}_1$

Probability $=\frac{3\times 4\times 2}{{}^9{C}_3}=\frac{24\times 3!}{9!}\times 6!=\frac{24\times 6}{9\times 8\times 7}=\frac{2}{7}$.