MarkFL said:
Yes, that's right about a bucket not getting a ball. I am looking at this from the perspective that each placement of a red ball is a separate event. I find it simpler that way. :)
So using what we have found so far, what is the probability that a certain bucket will get $n$ red balls, where $0\le n\le R$?
Let's look at a simple example...we have 3 red balls ($R=3$). Now suppose we want to compute the probability that a particular bucket gets 1 red ball. So, using what we have found, we would have:
$$\left(\frac{1}{M}\right)^1\left(\frac{M-1}{M}\right)^2$$
But...this is only the probability for one of the ways the bucket could get 1 red ball...it could get the red ball on the first distribution, or on the second, or on the third. So, the probability that a bucket will get 1 red ball is:
$$P(X=1)=3\left(\frac{1}{M}\right)^1\left(\frac{M-1}{M}\right)^2$$
Now, suppose we generalize back to $R$ red balls and $n$ red balls in a particular bucket. There will be $R$ distributions, and so we have $R$ "slots" to fill with either "Yes, got a ball" or "No, did not get a ball." Thus, we have $n$ yeses to distribute into $R$ slots.
For the first yes, we have $R$ choices, for the second we have $R-1$ choices and continuing all the way down to the $n$th yes, where we have $R-(n-1)$ choices. But, we must account for the fact that a yes going into a particular slot is equivalent to a yes going into that same slot on another choice. So, we must divide by the number of ways to arrange $n$ items, which is $n!$.
Putting all of this together, we find the number $N$ of distinct ways to distribute these yeses into the $R$ slots is given by:
$$N=\frac{R(R-1)(R-2)\cdots(R-(n-1))}{n!}=\frac{R!}{n!(R-n)!}={R \choose n}$$
What we have in fact just found is the number of ways to choose $n$ from $R$. :)
And so we find the probability that a particular bucket gets $n$ red balls is:
$$P(X=n)={R \choose n}\left(\frac{1}{M}\right)^n\left(\frac{M-1}{M}\right)^{R-n}$$
Now, if this is correct, then we should find that the sum of the probabilities for $n=0$ to $n=R$ is 1, since it is certain that any bucket will get some number of red balls from $0$ to $R$:
$$\sum_{n=0}^R\left({R \choose n}\left(\frac{1}{M}\right)^n\left(\frac{M-1}{M}\right)^{R-n}\right)=1$$
We should recognize that the LHS of the above equation is an application of the binomial theorem, and so we may write:
$$\left(\frac{1}{M}+\frac{M-1}{M}\right)^R=1$$
$$\left(\frac{M-1+1}{M}\right)^R=1$$
$$\left(\frac{M}{M}\right)^R=1$$
$$(1)^R=1$$
$$1=1$$
So, our formula checks out and we have in essence derived the binomial probability formula. Can you now use this formula to answer the given question:
$$P(X>1)=?$$