Suppose three numbers are randomly chosen from the following list: \begin{equation} 4,5,7,8,11 \end{equation} What is the probability that the numbers drawn represent sides of a triangle?
I posted my attempted solution below. Please let me know if it looks like a good strategy. It did take quite a while to figure out every possible combination of numbers which work, so if you think of a quicker approach please let me know. Thanks!
On
{4,5,11} and {4,7,11} can work too if you change the units of the "offending" side to force it to work so it depends on what you allow and what you disallow and since you didn't specify, I am saying 100% is possible cuz I can change the units for the 11 side to be something that completes the triangle.
My attempted solution:
Use the triangle inequality theorem: For any three side lengths $a,b,c$, these lengths represent sides of a triangle if $x+y \ge z$ holds for every possible $x,y,z \in \{a,b,c\}$.
Therefore, the following triples representing possible triangle side lengths are:
4,5,7
4,5,8
4,7,8
4,8,11
5,7,8
5,7,11
5,8,11
7,8,11
And the number of possibilities (where order does not matter) is: \begin{equation} _5C_3=\frac{5!}{3!2!}=\frac{120}{12}=10 \end{equation} Thus the desired probability is $\frac{8}{10}=\frac{4}{5}$.