Permutations and Combinations, Picking discs A bag contains 6 red, 5 blue and 4 green discs. If 3 discs are chosen at random, find the probability that there is at least one blue disc given that at least one red disc is chosen.
I thought that on the denominator is 6*14C2 as you pick one red and from the rest you pick 2. On the numerator you do 6*5*13, as you pick one red, one blue and one of the 13 rest.
This would give 5/7.
But the answer is 255/371.

A red disc is selected unless only blue and green discs are selected.  Therefore, the number of selections in which at least one red disc is selected is 
$$\binom{6 + 5 + 4}{3} - \binom{5 + 4}{3} = \binom{15}{3} - \binom{9}{3}$$
Selections that include both a blue disc and a red disc either contain one disc of each color, two red discs and a blue disc, or a red disc and two blue discs.  The number of such selections is 
$$\binom{6}{1}\binom{5}{1}\binom{4}{1} + \binom{6}{2}\binom{5}{1} + \binom{6}{1}\binom{5}{2}$$
Hence, the probability that a blue disc is selected given that at least one red disc is selected is 
$$\frac{\dbinom{6}{1}\dbinom{5}{1}\dbinom{4}{1} + \dbinom{6}{2}\dbinom{5}{1} + \dbinom{6}{1}\dbinom{5}{2}}{\dbinom{15}{3} - \dbinom{9}{3}}$$
