Question

Two dice appear to be standard 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 $m + n.$

Answer

Solution:

Since $1 + 2 + \cdots + 6 = 21,$ each die rolls $k$ with probability $\frac{k}{21}.$ A total of $7$ arises from the pairs $(k, 7-k)$ for $k = 1, \ldots, 6,$ so its probability is $\frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} = \frac{56}{441} = \frac{8}{63}.$

Thus $m + n = 8 + 63 = 71.$

2016 AIME I Problem 2

Solution:

Problem 2 in Other Years