# Probability of forming a triangle given sides to choose from

What is the probability of being able to form a triangle from three segments chosen at random from five line segments of lengths $1,3,5,7$ and $9$?

Denote with $A$ the event of forming a triangle from three segments chosen from the list above. Then, the segments \begin{aligned} &1)\,\,3-5-7\\ &2)\,\,5-7-9\\ &3)\,\,3-7-9\\ \end{aligned} are the only ones that can be used to create a triangle. Each case can be arranged in $3!$ ways, thus $N(A)=3!\cdot3$. On the other hand, $N=5\cdot4\cdot3\,$ and so $$P(A)=\frac{N(A)}{N}=\frac{3!\cdot3}{5\cdot4\cdot3}=0.3$$

Would this look right? I am assuming that a side can be chosen only once, if that's not the case, then the following wold be how I would do it:

The segments to choose from are now

\begin{aligned} &1)\,\,3-5-7\\ &2)\,\,5-7-9\\ &3)\,\,3-7-9\\ &4)\,\,1-1-1\\ &5)\,\,3-3-3\\ &6)\,\,5-5-5\\ &7)\,\,7-7-7\\ &8)\,\,9-9-9\\ \end{aligned} The first three cases can be arranged in $3!$ ways and the other five only in $1$ way. thus $N(A)=3!\cdot3 +5$. On the other hand, $N=5^3$ and so $$P(A)=\frac{N(A)}{N}=\frac{3!\cdot3+5}{5^3}=0.184$$

For these $2$ solutions I am not sure if the order matters or not. I know that the sides $3-5-7$ and $7-5-3$ form the same triangle, so would this mean that order does not matter? The way I argued is that if we set $A_1=3$ for the first and $A_1=7$ for the second, it is easy to see that $A_1\neq A_2$, so by this logic order matters.

• Order does not matter, we have to look at the possible combinations of size $3$ out of the possible $5$ segments. – Thomas Bladt Apr 21 '18 at 4:26
• If your repeat a segment then you are missing a bunch of them: 3-3-5, 5-5-7, 5-5-9, 7-7-9. Each of these can be picked $3$ ways so you need to add $3 \times 4 = 12$ to $N(A)$. – antkam Apr 21 '18 at 4:59

You can choose $3$ segments from those $5$ on ${5\choose 3} =10$ ways, but only $3$, you listed, are OK.- So the answer is $0.3$