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Probability - two unfair dice

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    The sum of the probabilities of each number on the dice is 1, i.e
    [tex]\frac{1}{6}+\frac{1}{6}+\frac{1}{9}+x+\frac{2}{9}+y=1[/tex]
    where x and y are the probabilities of number 4 and 6 respectively. Solving,
    [tex]x+y=\frac{1}{3}[/tex]

    The probability that the two dice shows same number is
    [tex]\left(\frac{1}{6} \right)^2+\left(\frac{1}{6} \right)^2+\left(\frac{1}{9} \right)^2+x^2+\left(\frac{2}{9} \right)^2+y^2=\left(\frac{2}{3} \right)^4[/tex]
    Solving, [tex]x^2+y^2=\frac{13}{18}[/tex]
    Rewriting ##x^2+y^2## as ##(x+y)^2-2xy## and substituting the value of ##x+y##,
    [tex]2xy=\frac{-11}{18}[/tex]
    For the sum of two resulting numbers to be 10, there are three possible cases. The first shows 4 and the second shows 6 or (4,6). The other cases are (5,5) and (6,4).
    The probability that the sum of the two resulting numbers is 10 can be given by the expression:
    [tex]2xy+\left(\frac{2}{9} \right)^2[/tex]
    Substituting the value of ##2xy##, I get a negative answer. :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

    Last edited: Mar 8, 2013
  2. jcsd
  3. Mar 8, 2013 #2

    Dick

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    I think you are just messing up the arithmetic. I don't get x^2+y^2=13/18. Check it.
     
  4. Mar 8, 2013 #3
    Oh yes, I missed a factor of 9 in the denominator. :redface:

    Thanks Dick! :smile:
     
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