# Difference between revisions of "2006 AMC 12B Problems/Problem 17"

## Problem

For a particular peculiar pair of dice, the probabilities of rolling $1$, $2$, $3$, $4$, $5$ and $6$ on each die are in the ratio $1:2:3:4:5:6$. What is the probability of rolling a total of $7$ on the two dice? $\mathrm{(A)}\ \frac 4{63} \qquad \mathrm{(B)}\ \frac 18 \qquad \mathrm{(C)}\ \frac 8{63} \qquad \mathrm{(D)}\ \frac 16 \qquad \mathrm{(E)}\ \frac 27$

## Solution

The probability of getting an $x$ on one of these dice is $\frac{x}{21}$.

The probability of getting $1$ on the first and $6$ on the second die is $\frac 1{21}\cdot\frac 6{21}$. Similarly we can express the probabilities for the other five ways how we can get a total $7$. (Note that we only need the first three, the other three are symmetric.)

Summing these, the probability of getting a total $7$ is: $$2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) = \frac{56}{441} = \boxed{\frac{8}{63}}$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 