- #1

- 230

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

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- Thread starter Somefantastik
- Start date

- #1

- 230

- 0

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

- #2

- 11

- 0

k*n/(m+n)

- #3

- 230

- 0

Thanks for your help, but I had the answer and was really looking for the process.

- #4

- 268

- 6

Look at the expectation of a Hypergeometric variable.

- #5

- 111

- 1

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

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 [itex]X_{j}[/itex] which equal 1 if the j-th drawn ball is blue and 0 if it is red. Now define the random variable

[tex]

X = \sum_{j=1}^{k} X_{j}

[/tex]

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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