# Mock AIME I 2015 Problems/Problem 4

## Problem 4

At the AoPS Carnival, there is a "Weighted Dice" game show. This game features two identical looking weighted 6 sided dice. For each integer $1\leq i\leq 6$, Die A has $\tfrac{i}{21}$ probability of rolling the number $i$, while Die B has a $\tfrac{7-i}{21}$ probability of rolling $i$. During one session, the host randomly chooses a die, rolls it twice, and announces that the sum of the numbers on the two rolls is $10$. Let $P$ be the probability that the die chosen was Die A. When $P$ is written as a fraction in lowest terms, find the sum of the numerator and denominator.

## Solution

The probability that A lands at 10 is $$\dfrac{4 \cdot 6 + 5 \cdot 5 + 6 \cdot 4}{441} = \dfrac{73}{441}.$$ The probability that B lands at 10 is $$\dfrac{1 \cdot 3 + 2 \cdot 2 + 3 \cdot 1}{441} = \dfrac{10}{441}.$$ The probability that A was picked is thus $\dfrac{73}{73+10} = \dfrac{73}{83}$. The sum of numerator and denominator is thus $\boxed{156}$.