Given a bag containing $8\ \color{red}{red}$ balls and $4\ \color{green}{green}$ balls,
what is the probability of drawing $3$ balls at random such that $\mathbf {none}$ of them are $\color{red}{red}$?
My Approach:
$$1-(\ P(Balls_{\color{red}{red}}\ge 1)+P(Balls_{\color{red}{red}}\ge 2)+P(Balls_{\color{red}{red}}\ge 3)\ )$$
$$1 – \left({8 \choose 1} {4 \choose 3} + {8 \choose 2} {4 \choose 1} + {8 \choose 3} {4 \choose 0}\right)= -423$$
$$Total\ possible\ ways={12 \choose 3}$$
Answer given:
$$\frac{4}{495}$$
I am getting the wrong answer and I don't know why. Please correct my work and tell me if there are any alternate approaches that I could use.
Best Answer
Directly, Pr = $\dfrac{4\choose3}{12\choose3} = \dfrac{1}{55}$
In the tortuous way you were attempting,
Pr = 1 - $\dfrac{{8\choose3}{4\choose0} +{8\choose2}{4\choose1} +{8\choose 1}{4\choose2}}{12\choose3} = \dfrac{1}{55} $