Two fair six-sided dice are tossed and the numbers shown on the top face of each are added together. What is the probability that the resulting sum is less than 10 ?
We call the numbers tossed on the two dice $A$ and $B$.

There are 6 possible values for each of $A$ and $B$, and each such value is equally likely.

Therefore, there are $6 \cdot 6=36$ equally likely possible pairs of outcomes.

If $A=1$, then for $A+B<10, B$ can equal any of $1,2,3,4,5,6$.

If $A=2$, then for $A+B<10, B$ can equal any of $1,2,3,4,5,6$.

If $A=3$, then for $A+B<10, B$ can equal any of $1,2,3,4,5,6$.

If $A=4$, then for $A+B<10, B$ can equal any of $1,2,3,4,5$.

If $A=5$, then for $A+B<10, B$ can equal any of $1,2,3,4$.

If $A=6$, then for $A+B<10, B$ can equal any of $1,2,3$.

Thus, $6+6+6+5+4+3=30$ pairs of outcomes satisfy the requirement and so the probability that $A+B$ is less than 10 is $\frac{30}{36}$ or $\frac{5}{6}$.

ANSWER: $\frac{5}{6}$