There are 3 different ways you could pick the balls: 1) Red, Red, Green vs. 2) Red, Green, Red vs. 3) Green, Red, Red.

Let's pick the first scenario. There are 7 out 12 Reds. Thus, the probability of picking a Red is 7/12. The probability of picking another Red is 6/11, since 6 Reds remain and there are now 11 total balls. The probability of picking a Green is 5/10.

\(\frac{7}{12} * \frac{6}{11} * \frac{5}{10}\)
\(= \frac{210}{1320}\)
\(= \frac{7}{44}\)

However, since there are 3 different ways, multiply by 3:
\(= \frac{7}{44}*3\)
\(= \frac{21}{44}\)