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)

$$\eqalign{ & {\text{Here}} \cr & n\left( E \right){ = ^7}{C_1}{ \times ^5}{C_1}{ \times ^5}{C_1} \cr & {\text{and,}} \cr & n\left( S \right){ = ^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \cr & P(S) = \frac{{7 \times 6 \times 5}}{{12 \times 11 \times 10}} \cr & = \frac{7}{{44}} \cr} $$