Probability of picking out balls of three different colours Problem

A bag contains $3$ red balls, $4$ green balls and $5$ blue balls. Compute the probability of picking out three balls of different colours, without replacement.

Possible solution
I compute the different combinations: RVB, RBV, VRB, VBR, BRV and BVR. Each one has probability $\frac{1}{22}$. Therefor I thought that the total probability was $6\cdot\frac{1}{22}$. Is my solution correct or completely wrong?
Thank you in advance and have a nice day.
Lorenzo Fabbri - San Marino

You are correct.  Those are disjoint probabilities and can be added.  They cover all the ways to get one of each color.
