Probability of at least 3 red balls given 4 choices in a bag of 4 red balls and 4 black balls? Let's say there are 8 balls in a bag, where 4 are red and 4 are black.
If I choose four balls from the bag without replacement, what is the probability that I will choose at least 3 red balls?
My thinking was to use the idea that $P(E) = \frac{|E|}{|S|}$. Therefore, am I correct in saying that $|E| = {4 \choose 3} \cdot 5$, since I am choosing 3 red balls from the 4 available, and the last ball can be of any colour?
However, I am not sure about $|S|$. How do I choose four balls from eight, keeping into account that there are only two colours? I assume that $8 \choose 4$ isn't correct.

Here we want to find the probability of only 3 Red balls OR 4 Red balls being drawn without replacement. This means that we need to add the probability of either event together.
Another way to think of it is finding the total number of ways there are to draw 3 Red balls/4 Red balls, and then divide that by the total number of possible outcomes.
the number of ways to choose 3 Red balls from 4 Red balls:   $$ 4 \choose 3 $$
The probability of that happening is equal to the probability that a Red is drawn in 3 of 4 slots (and by multiplication rule, we don't care how we arrange them) 
$$ \frac48 *\frac37*\frac26*\frac45$$
AND the 4th slot is Black. So we multiply by the $\frac 45$. 
the number of ways to draw 4 Red balls from 4 Red balls is 1: $${4\choose 4} = \frac{4!}{4!} = 1$$
Multiplied by the probability of pulling 4 Red balls which is: $$ \frac 48 
*\frac 37 * \frac 26 * \frac 15$$
Therefore we get :$$ {4 \choose 3} \frac48 *\frac37*\frac26*\frac45 + {4 \choose 4} \frac48 *\frac37*\frac26*\frac15 $$
$$= 17/70$$
