Thank You. That's nice.
But it's not what the original problem asks. It states and asks:
After having thrown R balls over M buckets, the probability that exactly n<R balls fall in one specific bucket is:
$$P(X=n)={R \choose n}\left(\frac{1}{M}\right)^n\left(1-\frac{1}{M}\right)^{R-n}$$
That...
Wow... I see. So, let's put the problem (the very 1st post of this thread) a bit differently.
Instead of asking "what is the probability to find n red balls into ANY bucket", I ask:
"What is the probability to find exactly n red balls into at least one bucket".
Here is the train of thoughts...
Hummm... Are you sure?
So, if I have M=3 buckets and I throw R=2 red balls above them the prob. of finding n=1 red ball in one specific bucket is (from the binomial formula):
P(X=1) = 0.444
Based on what you said, M x P(X=1) = 1.333
This is not a probability.
Greetings.
I would need help as follow up to the answers to my http://mathhelpboards.com/basic-probability-statistics-23/prob-red-ball-buckets-binomial-17282-post79735.html#post79735.
After having thrown R red balls over M buckets, the probability that exactly n red balls fall in one, specific...
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.
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?
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?
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...