My Math Forum Probability of selecting balls from buckets

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 May 13th, 2014, 02:43 AM #1 Newbie   Joined: May 2014 From: Germany Posts: 1 Thanks: 0 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?
 May 13th, 2014, 05:55 AM #2 Senior Member   Joined: Apr 2014 From: UK Posts: 919 Thanks: 331 Yes
 May 13th, 2014, 01:51 PM #3 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 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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### probability of balls

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