What is the probability of drawing 3 balls such that none of them is red? 
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.
 A: 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} $
A: The total no. of ways in which green balls can be chosen $={4\choose 3}$.
The total no. of ways in which balls can be chosen $={12\choose 3}$

Probability that the balls will not be red(i.e., they will be green)$= \frac{4\choose 3}{12\choose 3}=\frac{1}{55}$.

A: The probability of randomly drawing 3 balls not red would mean selecting green balls such that there is a combination of 4 green balls from which we can choose 3 divided by the combination of 12 balls, green and red, choosing 3 green balls at random therefore:

4C3 / 12C3 = 1/55 or 0.0182

A: Well you are summing events prob which are not independent, see there : https://en.wikipedia.org/wiki/Law_of_total_probability
and the approach given in comments by @true blue and using the very same theorem could lead you to the correct probability, with a three steps process, first draw : you get a green, same for second, etc, updating the odds with the right numbers of balls.
