I cannot fault your logic Naciema. That is because it is correct! Thank you for your answer. 
I think your method gives a solution of 34/455
I did it a different way and got a different answer the same answer.
I said that there are 4C3=4 ways to chose 3 red balls
There are 5C3=10 ways to chose 3 white
and 6C3=20 ways to choose 3 blue.
That is a total of 34 favourable outcomes
The sample space is 15C3 = 286 ways 455 ways (error corrected)
34/286 = 17/143 34/455
Just as Nacirema found
