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 August 27th, 2018, 05:37 AM #1 Newbie   Joined: Aug 2018 From: United States Posts: 1 Thanks: 0 Differing probabilities .. stuck need help I have tried to work this out but my answers never seem to agree with the actual computer program I wrote to do the work. Here is the scenario. You have 3 buckets. each bucket contains 10 marbles. In bucket A there are 2 white marbles and 8 black ones In bucket B there are 3 white marbles and 7 black ones In bucket C there are 4 white marbles and 6 black ones You randomly select one marble from each bucket. What are the odds of selecting AT LEAST 2 white marbles? Thanks in advance for any help ... want to learn how to do the require calculations, not just get the answer. Thanks again! August 27th, 2018, 09:05 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 At least two means exactly 2 or exactly 3. Let w = number of whites. $\text {P}(w = 2 \text { or } w = 3) = \text {P}(w = 2) + \text {P}(w = 3) - \text {P}(2 = w = 3).$ Now it is impossible for 2 = 3 $\implies \text {P}(2 = w = 3) = 0 \implies$ $\text {P}(w = 2 \text { or } w = 3) = \text {P}(w = 2) + \text {P}(w = 3) - 0 = \text {P}(w = 2) + \text {P}(w = 3).$ That is easy enough to figure out. $a = \text { probability of choosing white from bucket A } = \dfrac{2}{10} = \dfrac{1}{5}.$ $x = \text { probability of choosing black from bucket A } = 1 - \dfrac{1}{5} = \dfrac{4}{5}.$ $b = \text { probability of choosing white from bucket B } = \dfrac{3}{10}.$ $y = \text { probability of choosing black from bucket B } = 1 - \dfrac{3}{10} = \dfrac{7}{10}.$ $c = \text { probability of choosing white from bucket C } = \dfrac{4}{10} = \dfrac{2}{5}.$ $z = \text { probability of choosing black from bucket C } = 1 - \dfrac{2}{5} = \dfrac{3}{5}.$ There is only one way to get exactly 3 whites, namely to pick 1 white from each bucket. Are those events independent? Yes. So the probability of three whites $= abc = \dfrac{1 * 3 * 2}{5 * 10 * 5} = \dfrac{6}{250}.$ There are, however, six mutually exclusive ways to get exactly 2 whites. What are they? What are their probabilities? Now add it all up, and you are done. NOTE: This is not the most efficient way to compute the answer, but it is probably the most intuitive way for a beginner. Last edited by JeffM1; August 27th, 2018 at 09:08 AM. Tags differing, probabilities, stuck 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 MeXiCaN Probability and Statistics 4 November 15th, 2017 09:40 AM RifkiNada Algebra 2 November 24th, 2012 03:32 AM safyras Algebra 8 April 5th, 2011 11:48 AM carter22 Algebra 2 March 1st, 2010 05:49 PM Hugh_Compton Algebra 2 June 1st, 2009 02:46 PM

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