Expected number of blue balls drawn from a sack of m red balls and n blue balls?

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The discussion focuses on calculating the expected number of blue balls drawn from a sack containing m red balls and n blue balls when k balls are randomly selected. The process involves using the expectation of a Hypergeometric variable and translating the problem into mathematical terms using indicator random variables. The expected value of the total number of blue balls drawn is represented as the sum of these indicator variables. An important realization is that the answer can be factored out of the sum, simplifying the calculation. Understanding this process is crucial for accurately determining the expected outcome in such scenarios.
Somefantastik
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Can someone help me break this down?

\Sigma^{k}_{i=1}\frac{i \left(^{n}_{i}\right)\left(^{m}_{k-i}\right)}{\left(^{m+n}_{k}\right)}
 
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k*n/(m+n)
 
Thanks for your help, but I had the answer and was really looking for the process.
 
Look at the expectation of a Hypergeometric variable.
 
Somefantastik said:
Can someone help me break this down?

\Sigma^{k}_{i=1}\frac{i \left(^{n}_{i}\right)\left(^{m}_{k-i}\right)}{\left(^{m+n}_{k}\right)}


First translate from math to English: there are m red balls and n blue balls in a sack from which you randomly draw k balls. What is the expected number of blue balls drawn? Now translate back into math: try using indicator random variables X_{j} which equal 1 if the j-th drawn ball is blue and 0 if it is red. Now define the random variable

<br /> X = \sum_{j=1}^{k} X_{j}<br />

and compute the expected value of that and hopefully you'll get the answer that Roberto gave.


addendum: doh! After all that I just realized you can factor the answer out of the sum. Then use the illustration of selecting balls to see what the resulting sum must be.
 
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