MHB Prob of red ball in buckets. Binomial?

[paolopiace]

I know for many of you this sounds trivial. For me it's not...

A barrel contains M buckets each containing B balls, for a total of N = M x B balls.

N is a big number.

A small number R > 1 of balls are red. All the others N-R are white.

The location of each one of these R balls is random across the M buckets.
That means, one or more red balls can be in any bucket.

Now I randomly pull one bucket from the barrel.

What is the probability that X > 1 red balls are in that bucket?

Besides the final formula that I really need, may I also ask for a guidance to get to it?

Thank You!

 

[MarkFL]

I think we can ignore the white balls, and just imagine we have $M$ buckets before us, and $R$ red balls to randomly distribute among the buckets. Now, each time we put a red ball into a bucket, what is the probability that for any particular bucket, it will be the one that gets the ball?

 

[paolopiace]

... what is the probability that for any particular bucket, it will be the one that gets the ball?

I think there are R/M probabilities to see one red ball in a particular bucket.
Then, there are (R-1)/M probabilities to see a second red ball in that same bucket.
Therefore R(R-1)/M^2 is the prob. to find two red balls in a particular bucket.

Kind of right? Or not?

 

[MarkFL]

Each time we place a red ball into a bucket, the probability that a particular bucket will get the ball is:

$$P(X)=\frac{1}{M}$$

What is the probability that the bucket will not get the ball?

 

[paolopiace]

... What is the probability that the bucket will not get the ball?

It's 1-1/M

I am kind of confused. If a throw R balls in the air, there are R/M probabilities that a specific bucket gets a ball. Or not?

1/M is the probability that a specific bucket gets a ball if I throw one and only one ball. Or nor?

 

[MarkFL]

It's 1-1/M

I am kind of confused. If a throw R balls in the air, there are R/M probabilities that a specific bucket gets a ball. Or not?

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 $0\le n\le R$?

 

[MarkFL]

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:

$$P(X>1)=?$$

 

[MarkFL]

Just to finish this up, we have:

$$\sum_{n=0}^R\left({R \choose n}\left(\frac{1}{M}\right)^n\left(\frac{M-1}{M}\right)^{R-n}\right)=1$$

And we can write this as:

$${R \choose 0}\left(\frac{1}{M}\right)^0\left(\frac{M-1}{M}\right)^{R-0}+{R \choose 1}\left(\frac{1}{M}\right)^1\left(\frac{M-1}{M}\right)^{R-1}+\sum_{n=2}^R\left({R \choose n}\left(\frac{1}{M}\right)^n\left(\frac{M-1}{M}\right)^{R-n}\right)=1$$

Now, we should recognize that:

$$P(X>1)=\sum_{n=2}^R\left({R \choose n}\left(\frac{1}{M}\right)^n\left(\frac{M-1}{M}\right)^{R-n}\right)$$

And so, making that substitution, and simplifying, we obtain:

$$\left(\frac{M-1}{M}\right)^{R}+R\left(\frac{1}{M}\right)\left(\frac{M-1}{M}\right)^{R-1}+P(X>1)=1$$

Solving for $P(X>1)$, we find:

$$P(X>1)=1-\left(\frac{M-1}{M}\right)^{R}-\frac{R}{M}\left(\frac{M-1}{M}\right)^{R-1}$$

 

[paolopiace]

Just to finish this up, we have: ...

This is beautiful!
I want to thank you so much, although I still have to digest it entirely.
Sorry for this late feedback. Many things of life kept me away from my post.