My Math Forum Differing probabilities .. stuck need help

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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.

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