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.

  • $\begingroup$ Order does not matter, we have to look at the possible combinations of size $3$ out of the possible $5$ segments. $\endgroup$ – Thomas Bladt Apr 21 '18 at 4:26
  • $\begingroup$ 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 $N(A)$. $\endgroup$ – 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 $0.3$

Order is here irrelevant.

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  • $\begingroup$ Your answers assumes that order does not matter and that you cannot choose the same segment more than once. Why so? I'm more interested in the answer to this question, rather the numerical solution itself $\endgroup$ – DMH16 Apr 21 '18 at 4:26
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    $\begingroup$ If we suppose we have only one segment each (suppose some stick), then you can't use segments more then once. $\endgroup$ – Aqua Apr 21 '18 at 4:30
  • $\begingroup$ Thank you, the analogy made it clear. $\endgroup$ – DMH16 Apr 21 '18 at 4:31

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