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March 23rd, 2012, 07:27 AM  #1 
Newbie Joined: Mar 2012 Posts: 1 Thanks: 0  hello conceptual probability questions
hello new to here. I encounter a problem on probability and then I find out this forum on search engine. It's nice The problem is There are 3 blue ball and 1 red ball in a bag, one of the balls are drawn to another bag that contains 4 green ball. Then two balls are drawn successively from the bag with 4 green ball + one unknown colour ball (without replacement) What is the probability of getting two balls of the same colour. I thought there were 4 possiblities. GGGB, GGGB, GGGB and GGGR (G: Green B:Blue R:Red) for each possibe situation I set up a formula (3/4) x (2/3) then for four possible situation, I then sum up to get (3/4) x (2/3) + (3/4) x (2/3) + (3/4) x (2/3) + (3/4) x (2/3) but the result is undoubtedly wrong because it's bigger than 1 which is impossible. then the only solution is (3/4) x (2/3) from my guess. But why..... Why I don't need to split into 4 situation? 
March 23rd, 2012, 03:36 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,755 Thanks: 695  Re: hello conceptual probability questions
The straightforward approach is as follows. To get 2 balls of the same color, they both must be green. The probability that the first ball is green is 4/5. Assuming the first ball is green, the probability the second ball is green is 3/4. Therefore the probability that both are green is 4/5 x 3/4 = 3/5.


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