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.
Best Answer
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 $\binom{8}{4}$.