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Old 03-10-2017, 02:32 PM   #1
DMMx69
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Probability choosing combo of red/black balls from a pool, no replacement?

I'm sure this is a simple solution, but it's not coming to me.

I have a bag of 5 black balls and 3 red balls.

I pull out 4 balls: 3 red and 1 black. I am NOT putting the pulled-out ball back in before the next choice.

What is the chance of pulling various combos of red/black out, NOT replacing the ball each time? How can I generalize this problem?

Thank you in advance.
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Old 03-10-2017, 06:09 PM   #2
heehaww
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Re: Probability choosing combo of red/black balls from a pool, no replacement?

P(x red and y black) = C(3,x) * C(5,y) / C(8, x+y)

It's just like if you were picking red/black cards from a deck.
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Old 03-13-2017, 09:40 PM   #3
Siegmund
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Re: Probability choosing combo of red/black balls from a pool, no replacement?

In general the answer to "I pull n balls from a bag with B black balls and R red balls; how many are black?" is a Hypergeometric distribution.. In Wikipedia's notation, N=total number of balls, K=number of red balls, n=number of balls you draw, k=number of red balls you get.

For your specific problem, as heehaww said,
4 black: 10 ways out of 56
3 black 1 red: 30 ways out of 56
2 black 2 red: 15 ways out of 56
1 black 3 red: 1 way out of 56
all 4 red obvious impossible.
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