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What is the probability that a random sample of 10 balls, taken from a collection of 50 blue balls + 50 red balls, has 5 blue balls and 5 red balls? When the balls are removed from the collection, they are not replaced.

I believe I will need some formula to account for all the orders in which 5 red balls and 5 blue balls can be collected. I think the probability that 1 red ball, then 1 blue ball, then one red ball, then one blue ball, ... is taken thus resulting in one example of how 5 red and 5 blue balls could be taken would be:

(50/100)(50/99)(49/98)(49/97)(48/96)(48/95)(47/94)(47/93)(46/92)*(46/91) = .001029

So then I am looking for some way to calculate the probability for all of the different orders that 5 red and 5 blue balls can be taken, and then I can add these together. This is where I am getting stuck. I could do it all by hand but what I want is an equation. This way it could easily work for different ratios of blue to red balls, different total number of balls, and also potentially include a third color of ball. Any help would be greatly appreciated!

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    $\begingroup$ Such a sample can be described as a sample of five red out of 50 and five blue out of 50 blue balls, whence there are $\binom{50}{5}\times\binom{50}{5}$ of them. The total number of ten-ball samples is $\binom{50+50}{5+5}.$ Divide the first by the second to obtain the probability when all samples are equally likely. Your question is a special case of the question asked and answered at stats.stackexchange.com/questions/37760. $\endgroup$
    – whuber
    Apr 30, 2022 at 21:07

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This can be described with a hypergeometric distribution.

See for instance the description on Wikipedia

the hypergeometric distribution is a discrete probability distribution that describes the probability of $k$ successes (random draws for which the object drawn has a specified feature) in $n$ draws, without replacement, from a finite population of size $N$ that contains exactly $K$ objects with that feature, wherein each draw is either a success or a failure.

In your case $k=5, n=10, K= 50, N = 100$.

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