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

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 colours is (A) (2/7) (B) (1/21) (C) (2/23) (D) (1/3)

Tardigrade Admin · Accepted Answer

n(S) = 9C3 n(E) = 3C1× 4C1 ×2C1 Probability = (3×4 ×2/9C3) = (24 ×3!/9!) ×6! = (24 ×6/9 ×8 ×7) = (2/7)

Q. 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

$\frac{2}{7}$

$\frac{1}{21}$

$\frac{2}{23}$

$\frac{1}{3}$

$n (S) =^{9} C_{3}$
$n (E) =^{3} C_{1} \times^{4} C_{1} \times^{2} C_{1}$
Probability $= \frac{3 \times 4 \times 2}{^{9} C _{3}} = \frac{24 \times 3 !}{9 !} \times 6! = \frac{24 \times 6}{9 \times 8 \times 7} = \frac{2}{7}$

Q. 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

72​

211​

232​

31​

n(S)=9C3​ n(E)=3C1​×4C1​×2C1​ Probability =9C3​3×4×2​=9!24×3!​×6!=9×8×724×6​=72​

$\frac{2}{7}$

$\frac{1}{21}$

$\frac{2}{23}$

$\frac{1}{3}$

$n (S) =^{9} C_{3}$
$n (E) =^{3} C_{1} \times^{4} C_{1} \times^{2} C_{1}$
Probability $= \frac{3 \times 4 \times 2}{^{9} C _{3}} = \frac{24 \times 3 !}{9 !} \times 6! = \frac{24 \times 6}{9 \times 8 \times 7} = \frac{2}{7}$