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
A) $\frac{1}{3}$
B) $\frac{2}{7}$
C) $\frac{1}{21}$
D) $\frac{2}{23}$
Explanation
(b)
Total number of cases$= \ ^{9}C_{3} = 84$
Number of favourable cases$= \ ^{3}C_{1}{.\ }^{4}C_{1}{.\ }^{2}C_{1} = 24$
$\therefore\ P = \frac{24}{84} = \frac{2}{7}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.