Probabilities: Estimating the probability of overlapping

  • Thread starter jksacc
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Main Question or Discussion Point

Hi all
I recently ran into this problem:
I have two bins. Each bin contains N numbered balls, from 1 to N.
For both the bins, the probability of the ball numbered k to be
selected equals to P(ball-k-selected)=k/SUM(1:N) (in other versions
this can be any given probability distribution)

Simple case:
Having selected 1 ball from the first bin, and 1 ball from
the second bin, i want to find the probability of the ball
having the same number.

If i am correct, the probability for this is SUM(k=1:N) (P(ball-k-selected)^2).

Complex case:
Having selected m balls from the first bin, and m balls from
the second bin, i need the probability of holding at least
one pair of balls with the same number at the end of the
experiment.

Assumption: The selection is without replacement. However, for
simplicity we can assume that the probability of a ball to be selected
remains stable during the experiment, and is given by
P(ball-k-selected)=k/SUM(1:N)

If something is not clear, please let me know.

Thanks in advance for any contributions!
 

Answers and Replies

  • #2
EnumaElish
Science Advisor
Homework Helper
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Denote integers with lowercase and sets with UPPERCASE. The set of balls in bin i is Ni.

There are C(n, m) = n!/(m!(n-m)!) combinations (subsets) that you can draw from either bin.

Let S1 be a subset of m balls from the 1st bin. Let S2 be the corresponding subset from the 2nd bin. How many subsets of m balls can you form out of the set of the remaining balls in the 2nd bin, S2' = N2\S2? The answer is s2' = |S2'| = C(n-m, m). That's the answer to the question, "for a given S1, how many disjoint subsets of the same size are there?" Since there are C(n,m) ways to construct S1, there are C(n,m)C(n-m,m) ways to construct two disjoint subsets, each with m elements.

Now you need to calculate the probability of obtaining these disjoint subsets.
 
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