Question:
Dakota randomly selected three different integers $1$ through $6.$ What is the probability that the three numbers selected could be the sides of a triangle? Express your answer as a common fraction.

Answer:
There are $\binom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20$ possible sets of three different integers. We need to figure out how many of these could be the sides of a triangle.

Clearly, none of the sides can be $1,$ since that would violate the Triangle Inequality. As for the rest, it is quite a simple matter to just list them all in an organized fashion: \begin{align*}
&(2, 3, 4)\\
&(2, 4, 5)\\
&(2, 5, 6)\\
&(3, 4, 5)\\
&(3, 4, 6)\\
&(3, 5, 6)\\
&(4, 5, 6)
\end{align*} Therefore, there are $7$ possible sets of numbers that could be sides of a triangle, out of $20$ possible sets, so our answer is $\boxed{\frac{7}{20}}.$