# Probability - two unfair dice

(see attachment)

## 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!

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Dick