Probability - two unfair dice

  • Thread starter Saitama
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  • #1
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Homework Statement


(see attachment)


Homework Equations





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!
 

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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
  • #3
3,816
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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.

Oh yes, I missed a factor of 9 in the denominator. :redface:

Thanks Dick! :smile:
 

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