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May 13th, 2014, 02:43 AM   #1
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Probability of selecting balls from buckets

Let's say we have one bucket with n red balls and m green balls.
I will select two balls from the bucket but with two different ways. Either one ball at a time or graping them together with one effort.
Is the probability of having one red and one green ball the same for the two cases?
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May 13th, 2014, 05:55 AM   #2
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Yes
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May 13th, 2014, 01:51 PM   #3
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Math Focus: Geometry
It might not be.
1. One ball at a time Order matters.
Red-Red = $\displaystyle \frac{n(n-1)}{n+m(n+m-1)}$

Red-Green = Green-Red = $\displaystyle \frac{mn}{(n+m)(n+m-1)}$

Green-Green = $\displaystyle \frac{m(m-1)}{(n+m)(n+m-1)}$

2. Two Together
Red-Red = $\displaystyle \frac{n^2}{(n+m)^2}$

Green-Green = $\displaystyle \frac{m^2}{(n+m)^2}$

Red-Green = $\displaystyle \frac{2mn}{(n+m)^2}$
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