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.
 A: Actually, $\binom{8}{4}$ is the correct value for $|S|$ (the way you're evaluating the probability, at least), but your value for $|E|$ is a bit off.
To see why your answer for $|E|$ is wrong, note that you're overcounting some selections several times. In particular, you're overcounting the case when you select all four red balls; for example, labeling the red balls $r_1, r_2, r_3, r_4$ and the black balls $b_1, b_2, b_3, b_4$, you count the selection $\{r_1, r_2, r_3, r_4\}$ four times; once when you choose $\{r_1, r_2, r_3\}$ as your first three red balls and $r_4$ as the fourth ball, once when you choose $\{r_1, r_2, r_4\}$ as your first three red balls, and $r_3$ as the fourth ball, etc.
The correct value for $|E|$ should be $\binom{4}{3}\cdot 4 + \binom{4}{4} = 17$; that is, the number of ways to choose 3 red balls and one black ball, plus the number of ways to choose all 4 red balls.
On the other hand, $|S|$ is indeed $\binom{8}{4}$. Since we labeled the balls, it doesn't matter that the balls are colored; the number we want is just the number of ways to choose 4 elements from a set of 8, which is just $\binom{8}{4}$.
A: Your method counts ways to select three red balls into one bag any ball into another.  Suppose someone secretly marked one of the red balls; then it could be one of the three red balls selected and one of the other three could be selected among the 5 other balls; or all three other red balls could be selected and the marked ball could be the one selected from the five other balls.  However these are the same thing.
To avoid double counting the favoured case you should count ways to pick either exactly 3 or exactly 4 red balls.
$$|E| = {4\choose 3}{4\choose 1}+{4\choose 4} = 17$$
The universal case is the way to choose any 4 balls from 8: $|S|= {8\choose 4}$
Remark The universal case and the favoured case should always count outcomes in the same manner; the favoured case has special considerations, and the universal is the most general form.  Here we are counting the ways to select 4 balls from 8, with special consideration for colours in the favoured case.
A: It depends if you can see the color of the balls when you choose them and other factors as well. Your original question did say chosen at random.
You also didn't state if the balls are all equally likely to be drawn.  It could be that the red ones are much smaller than the black ones or vice versa.
If the ball choices are at random and the balls are not replaced, then I get $17/70$ probability which is about $24.3$%.  The numerator comes from
$$\binom{4}{3}\binom{4}{1} + \binom{4}{4}\binom{4}{0}$$ 
and the denominator comes from
$$\binom{8}{4}$$
A: 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 :$$ {4 \choose 3} \frac48 *\frac37*\frac26*\frac45 + {4 \choose 4} \frac48 *\frac37*\frac26*\frac15 $$
$$= 17/70$$
