Find the probability of getting 2 golden coins from a bag of 4 golden coins and 8 iron coins with the following conditions: There are 4 golden coins and 8 iron coins in a bag. You select one coin from the bag, if it is a golden coin, you keep it; but if it is an iron coin, you put it back in the bag. Find the probability of earning exactly 2 golden coins after three consecutive attempt.

My Try:
Golden Coins=4
Iron coins = 8
Total coins = 12
We have 3 cases here: (GIG, GGI, IGG)
$ (\frac{4}{12}\times \frac{8}{11}\times \frac{3}{11})+(\frac{4}{12}\times \frac{3}{11}\times \frac{8}{10})+(\frac{8}{12}\times \frac{4}{12}\times \frac{3}{11}) = \frac{362}{1815} $
Is this answer correct?

We have 3 cases here: (GIG, GGI, IGG)
Every time that we take out a golden ball(G), we subtract 1 from the total balls.
Every time we take out an iron ball(I), we don't subtract anything because the ball goes back into the bag.
So, for  (GIG, GGI, IGG), we have:
$ (\frac{4}{12}\times \frac{8}{11}\times \frac{3}{11})+(\frac{4}{12}\times \frac{3}{11}\times \frac{8}{10})+(\frac{8}{12}\times \frac{4}{12}\times \frac{3}{11}) = \frac{362}{1815} $
