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 :
Solutions · 1
Official solution
Out of nine balls three balls can be chosen = 9C3 ways
\[\therefore\] Sample space = 9C3 = \[{{9!} \over {3!6!}}\] = \[{{9 \times 8 \times 7} \over 6}\] = 84
According to the question, all three ball should be different. So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen.
\[\therefore\] Total cases = \[{}^3{C_1} \times {}^4{C_1} \times {}^2{C_1}\] = 3 \[ \times \] 4 \[ \times \] 2 = 24
\[\therefore\] Probability = \[{{24} \over {84}}\] = \[{{2} \over {7}}\]