We have $6$ segments with the following lengths: $1,2,6,9,10$ and $11$. What is the probability of forming a triangle when choosing $3$ segments?
$3$ segments form a triangle if and only if the length of each one is less than the sum of others' length. The count of all possible variations is $C_6^3=\dfrac{6.5.4}{2.3}=20$. Now I am trying to count the variations which form a triangle. The only way that I found is to write all possible outcomes and see which do form a triangle $(6)$. Is there any better way to count it?
Sort it. In your case it is sorted already $1,2,6,9,10,11$
Pick ending $11$ and start from $1$ and $10$. If it were $1+10>11$, we would have a triangle. It is not.
Pick the next after $1$. It is $2$. Try again $2+10>11$. We have a triangle, and this means that besides that one triangle all combinations in between $6+10$, $9+10$ form a triangle as well. That is $3$ triangles.
Since we have a triangle now pick the one before $10$, and that is $9$ and try again. $2+9>11$. It is not a triangle so pick the one after $2$ and that is $6$. $6+9>11$ and that is a triangle and since there is no number between $6$ and $9$ that is the only additional triangle.
With that we exhausted all triangles that may have the largest side $11$. Move to the lower one $10$ and repeat.
In general you have this scheme:
$$ \begin{bmatrix} \mathbf{1} & 2 & 6 & 9 & \mathbf{10} & 11 & & \\ 1 & \mathbf{2} & 6 & 9 & \mathbf{10} & 11 & \text{+} & 3\\ 1 & \mathbf{2} & 6 & \mathbf{9} & 10 & 11 & & \\ 1 & 2 & \mathbf{6} & \mathbf{9} & 10 & 11 & \text{+} & 1\\ \mathbf{1} & 2 & 6 & \mathbf{9} & 10 & & & \\ 1 & \mathbf{2} & 6 & \mathbf{9} & 10 & & \text{+} & 2\\ 1 & \mathbf{2} & \mathbf{6} & 9 & 10 & & & \\ \mathbf{1} & 2 & \mathbf{6} & 9 & & & & \\ 1 & \mathbf{2} & \mathbf{6} & 9 & & & & \\ \mathbf{1} & \mathbf{2} & 6 & & & & & \\ \end{bmatrix} $$
That is making it $6$ out of $\binom{6}{3}=20$ and that is a probability of $\frac{3}{10}$