AMC 12 Mock Exam 5 Problem 8

The roll $(x,y,z)$ fails exactly when one number is at least the sum of the other two. At most one number can have this property (if $x\ge y+z$, then $x>y$ and $x>z$, so neither $y$ nor $z$ can be at least the sum of the other two). So the number of failing rolls is $3$ times the number of rolls with $x\ge y+z$.

For $y+z=s$ (where $2\le s\le6$) there are $s-1$ pairs $(y,z)$ and $7-s$ choices of $x\in\{s,\dots,6\}$, giving $$\sum_{s=2}^{6}(s-1)(7-s)=5+8+9+8+5=35.$$ Hence $3\cdot35=105$ of the $216$ rolls fail, and the probability of a triangle is $\frac{216-105}{216}=\frac{111}{216}=\boxed{\tfrac{37}{72}}$.