Probability - two unfair dice

1. Mar 8, 2013

Saitama

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
$$\frac{1}{6}+\frac{1}{6}+\frac{1}{9}+x+\frac{2}{9}+y=1$$
where x and y are the probabilities of number 4 and 6 respectively. Solving,
$$x+y=\frac{1}{3}$$

The probability that the two dice shows same number is
$$\left(\frac{1}{6} \right)^2+\left(\frac{1}{6} \right)^2+\left(\frac{1}{9} \right)^2+x^2+\left(\frac{2}{9} \right)^2+y^2=\left(\frac{2}{3} \right)^4$$
Solving, $$x^2+y^2=\frac{13}{18}$$
Rewriting $x^2+y^2$ as $(x+y)^2-2xy$ and substituting the value of $x+y$,
$$2xy=\frac{-11}{18}$$
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:
$$2xy+\left(\frac{2}{9} \right)^2$$
Substituting the value of $2xy$, I get a negative answer.

Any help is appreciated. Thanks!

Attached Files:

• probability.png
File size:
64.2 KB
Views:
81
Last edited: Mar 8, 2013
2. Mar 8, 2013

Dick

I think you are just messing up the arithmetic. I don't get x^2+y^2=13/18. Check it.

3. Mar 8, 2013

Saitama

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

Thanks Dick!