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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. RedRed = $\displaystyle \frac{n(n1)}{n+m(n+m1)}$ RedGreen = GreenRed = $\displaystyle \frac{mn}{(n+m)(n+m1)}$ GreenGreen = $\displaystyle \frac{m(m1)}{(n+m)(n+m1)}$ 2. Two Together RedRed = $\displaystyle \frac{n^2}{(n+m)^2}$ GreenGreen = $\displaystyle \frac{m^2}{(n+m)^2}$ RedGreen = $\displaystyle \frac{2mn}{(n+m)^2}$ 

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