Balls of 3 colours in a bag - Probability There are $12$ balls in a bag. $3$ of them are red, $4$ of them are green, and $5$ of them are blue. We randomly take out $3$ balls from the bag at the same time. What is the probability that all three balls are of the same colour?

My answer: $(3/12)^3 + (4/12)^3 + (5/12)^3$. Is this correct?

EDIT: My explanation is that since there are the odds of getting a red ball are 3/12, I simply multiplied it 3 times since I consider multiplication to be the "and" operator. Similarly, I consider addition to be the "or" operator. So, the way I think of it is: 

(1 red and 1 red and 1 red) or (1 green and 1 green and 1 green) or (1 blue and 1 blue and 1 blue).
There are  ${5\choose{3}} + {4 \choose 3} + {3 \choose 3}$ total ways of having a success.  There are a total of $12 \choose 3$ different ways of choosing 3 balls.

So the answer is $$\dfrac{{5\choose{3}} + {4 \choose 3} + {3 \choose 3}}{12 \choose 3} = \dfrac{3}{44} $$
Combination ($\binom n r$) can only be used when we can differentiate the balls even when they have the same colour, but if they are all of a particular colour they cannot be distinguished and hence the number of ways to pick up $r$ balls from a set of $n$ balls of a particular colour is only $1$ and not $binom n r$, and the probability  is $\dfrac r n$.