2016 AIME I Problems/Problem 2

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

Solution

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 rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\frac{56}{441}=\frac{8}{63}$ . The answer is $8+63=\boxed{071}$

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2016 AIME I Problems/Problem 2

Problem 2

Solution

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