A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:
A. $$\frac{{343}}{{1728}}$$
B. $$\frac{{21}}{{36}}$$
C. $$\frac{{12}}{{35}}$$
D. $$\frac{{7}}{{44}}$$
Answer: Option D
Solution (By Examveda Team)
$$\eqalign{ & {\text{Here}} \cr & n\left( E \right){ = ^7}{C_1}{ \times ^5}{C_1}{ \times ^5}{C_1} \cr & {\text{and,}} \cr & n\left( S \right){ = ^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \cr & P(S) = \frac{{7 \times 6 \times 5}}{{12 \times 11 \times 10}} \cr & = \frac{7}{{44}} \cr} $$This Question Belongs to Arithmetic Ability >> Probability
Join The Discussion