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!
Best Answer
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 $\frac{8}{10}=\frac{4}{5}$.