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 
, Die A has 
probability of rolling the number 
, while Die B has a 
probability of rolling . 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 
. Let 
be the probability that the die chosen was Die A. When 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 ![\[\dfrac{4 \cdot 6 + 5 \cdot 5 + 6 \cdot 4}{441} = \dfrac{73}{441}.\]](http://latex.artofproblemsolving.com/3/3/3/333d52bd045b69adf554aa1b1e23cf9b6c625bfb.png)
The probability that B lands at 10 is ![\[\dfrac{1 \cdot 3 + 2 \cdot 2 + 3 \cdot 1}{441} = \dfrac{10}{441}.\]](http://latex.artofproblemsolving.com/c/c/e/ccef5cc36ebe94a41e5d7541fae26ad0006bc3c8.png)
The probability that A was picked is thus 
. The sum of numerator and denominator is thus 
.
