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: 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 
$$\frac{\dbinom{6}{1}\dbinom{5}{1}\dbinom{4}{1} + \dbinom{6}{2}\dbinom{5}{1} + \dbinom{6}{1}\dbinom{5}{2}}{\dbinom{15}{3} - \dbinom{9}{3}}$$
A: *

*$\Pr\left(\text{no red}\right)=\frac{9}{15}\frac{8}{14}\frac{7}{13}=\frac{504}{2730}$

*$\Pr\left(\text{no blue}\right)=\frac{10}{15}\frac{9}{14}\frac{8}{13}=\frac{720}{2730}$

*$\Pr\left(\text{green}\right)=\frac{4}{15}\frac{3}{14}\frac{2}{13}=\frac{24}{2730}$ 


Substituting this in:
$$\Pr\left(\text{no blue}\right)=$$$$\Pr\left(\text{no blue}\mid\text{at least one red}\right)\Pr\left(\text{at least one red}\right)+\Pr\left(\text{no blue}\mid\text{no red}\right)\Pr\left(\text{no red}\right)=$$$$\Pr\left(\text{no blue}\mid\text{at least one red}\right)\left(1-\Pr\left(\text{no red}\right)\right)+\Pr\left(\text{green}\right)$$
we find:
$\frac{720}{2730}=\Pr\left(\text{no blue}\mid\text{at least one red}\right)\left(1-\frac{504}{2730}\right)+\frac{24}{2730}$
Leading to $\Pr\left(\text{no blue}\mid\text{at least one red}\right)=\frac{696}{2226}$
and consequently:
$\Pr\left(\text{at least one blue}\mid\text{at least one red}\right)=1-\Pr\left(\text{no blue}\mid\text{at least one red}\right)=\frac{1530}{2226}=\frac{255}{371}$
