Q. Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls is
A) $\frac{7}{\ ^{11}C_{7}}$
B) $\frac{\ ^{5}C_{3} + \ ^{6}C_{4}}{\ ^{11}C_{7}}$
C) $\frac{\ ^{5}C_{2}\ ^{6}C_{2}}{\ ^{11}C_{7}}$
D) $\frac{\ ^{6}C_{3}\ ^{5}C_{4}}{\ ^{11}C_{7}}$
Explanation
(c)
Favourable ways,
$= \ ^{5}C_{3}{\ \times}^{6}C_{4} = \ ^{5}C_{2}{\times \ }^{6}C_{2}$
and total number of ways$= \ ^{11}C_{7}$
$\therefore$Required probability$= \frac{\ ^{5}C_{2}{\times \ }^{6}C_{2}}{\ ^{11}C_{7}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.