2016 AIME I Problems/Problem 2

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Problem 2

Two dice appear to be normal dice with their faces numbered from $1$ to $6$, but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$. The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


It is easier to think of the dice as 21 sided dice with 6 sixes, 5 fives, etc. Then there are 21^2=441 possible roles. There are 2*(1*6+2*5+3*4)=56 roles that will result in a seven. The odds are therefore $56/441=8/63$. The answer is $8+63=71$

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