JEE MAIN - Mathematics (2010 - No. 5)
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 :
$${2 \over 7}$$
$${1 \over 21}$$
$${1 \over 23}$$
$${1 \over 3}$$
Explanation
Out of nine balls three balls can be chosen = $${}^9{C_3}$$ ways
$$\therefore$$ Sample space = $${}^9{C_3}$$ = $${{9!} \over {3!6!}}$$ = $${{9 \times 8 \times 7} \over 6}$$ = 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 = $${{24} \over {84}}$$ = $${{2} \over {7}}$$
$$\therefore$$ Sample space = $${}^9{C_3}$$ = $${{9!} \over {3!6!}}$$ = $${{9 \times 8 \times 7} \over 6}$$ = 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 = $${{24} \over {84}}$$ = $${{2} \over {7}}$$
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