QJEEM10Mad438: JEE Main 2010 Mathematics - Class 12 JEEM10Mad438 Solution: An urn contains nine balls of which three are red, four are…

Out of nine balls three balls can be chosen = $${}^9{C}_3$$ ways

$$\therefore$$ Sample space = $${}^9{C}_3$$ = $$\frac{9!}{3!6!}$$ = $$\frac{9\times 8\times 7}{6}$$ = 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 = $$\frac{24}{84}$$ = $$\frac{2}{7}$$