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.
On
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:
$\Pr\left(\text{at least one blue}\mid\text{at least one red}\right)=1-\Pr\left(\text{no blue}\mid\text{at least one red}\right)=\frac{1530}{2226}=\frac{255}{371}$
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 $$\frac{\dbinom{6}{1}\dbinom{5}{1}\dbinom{4}{1} + \dbinom{6}{2}\dbinom{5}{1} + \dbinom{6}{1}\dbinom{5}{2}}{\dbinom{15}{3} - \dbinom{9}{3}}$$