# Difference between revisions of "1996 AHSME Problems/Problem 16"

## Problem

A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?

$\text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{91}{216}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{8}{15}\qquad\text{(E)}\ \frac{7}{12}$

## Solution

The third die cannot be $1$, since the minimal sum on the other two dice is $2$.

If the third die is $2$, then the first two dice must be $(1,1)$.

If the third die is $3$, then the first two dice must be $(1,2)$ or $(2,1)$.

If the third die is $4$, then the first two dice must be $(1,3)$, $(2,2)$, or $(3,1)$.

If the third die is $5$, then the first two dice must be $(1,4)$, $(2,3)$, $(3,2)$, or $(4,1)$.

If the third die is $6$, then the first two dice must be $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, or $(5,1)$.

There are $15$ possibilities for the three dice. Of those possibiltiies, $7$ have a $2$ in the first two dice, and $1$ has a $2$ in the third die. Therefore, the answer is $\boxed{\frac{8}{15}}$.