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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: 6,806 Thanks: 716 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. Tags conceptual, probability, questions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ungeheuer Applied Math 0 February 26th, 2014 05:41 AM guru123 Algebra 3 August 26th, 2013 06:57 AM dragonaut Algebra 0 March 27th, 2013 12:46 PM nicoleb Calculus 7 November 12th, 2012 11:01 AM LastXdeth Calculus 4 February 13th, 2012 07:43 PM

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