I am reviewing this problem, say there are n balls in the urn, m of them is red, n - m of them is not red. Now, draw k balls out of the n balls without replacement, what is the probability of getting i red balls out of it.
So I watched a solution saying that let event A be the number of ways to draw i red balls from a total of m, and let the event B be the event that k-i non-red balls drawn from n-m non-red balls. So the event of ways of drawing i red balls out of total k balls is: $$P(C) = C(m, i)*C(n-m, k-i) / C(n, k)$$ While, I have no problem understanding the calculation of total sample size, which is simply n chooses k, I do not see why we could multiply the event A and event B, because it seems we are drawing separately assuming we know which part of the balls in urn is exactly red, and which is non-red. Could someone please explain the logic to me? Thank you!
After reading SmileyCraft's post, I drew a simple example to understand this problem. If we have a total of 9 balls, 5 of them are numbered, 4 of them are labeled with alphabets, then it is not hard to imagine an urn with the following balls, ball is within parenthesis. (1), (2), (3), (4), (5),| (a), (b), (c), (d) Suppose we draw a total of 3 balls, and we want to know the probability of 2 of them is numbered balls. If we apply the same logic as it is in the question I mentioned: Step 1. Choose 2 balls out of the 5 numbered balls Step 2. Choose 1 ball out of the 4 alphabet labeled balls
The total sample space $\Omega$ is C(9, 3)
The step of C(5,2), which is in Step 1: there are the following combinations: (1)(2),(1)(3),(1)(4),(1)(5),(2)(3),(2)(4),(2)(5),(3)(4),(3)(5),(4)(5) A total of 10 combinations, same as C(5,2) = 5!/(2!*3!)
The step of C(4,1), is the Step 2: (a), (b), (c), (d). 4 different combinations = C(4,1).
Now, the logic is that, the reason that they can be multiplied is because, each of the alphabet combination can be paired with each of the numbered combination such as:
For (a) combination to be paired with numbered:
Total: 1x10: (a)(1)(2), (a)(1)(3),...,(a)(2)(3),...,(a)(4)(5)
and exhaust all the combinations in the 2 numbered balls.
Same for (b) combination paired with those in Step 1:
Total: 1x10:(b)(1)(2), (b)(1)(3),...,(b)(2)(3),...,(b)(4)(5)
Similarly, we have
Total: 1x10:(c)(1)(2), (c)(1)(3),...,(c)(2)(3),...,(c)(4)(5)
Total: 1x10:(d)(1)(2), (d)(1)(3),...,(d)(2)(3),...,(d)(4)(5)
Now, we have 10*4 (or 4*10) = 40 combinations of choosing 3 balls with exactly 2 numbered balls. This 10*4 is exactly C(5,2)*C(4,1)
And this can be extend to different n, m, n-m, k, and i.
I hope this is correct, please kindly comment on it.
I felt it was more easier to see this process with the labels on, then looking at this problem without the labels so no clues about which ball is which. But imagine we take off the labels, it's still the same process.
We simply need to know the amount of ways to pick $k$ balls such that $i$ are red. Equivalently, ways to pick $i$ red balls and $k-i$ non-red balls. For every such way we have a corresponding way to choose $i$ balls out of $m$ and a corresponding way to choose $k-i$ balls out of $n-m$. The important thing is that this goes both ways. For any combination of a way to choose $i$ balls out of $m$ and a way to choose $k-i$ balls out of $n-m$, there is a corresponding way to choose $i$ red balls and $k-i$ non-red balls.