# A sticky problem of probability

A box contains seven sticks which are respectively 2cm,3cm,4cm,7cm,8cm,and 11cm long.

How do I find the probability that any three sticks chosen randomly from the box to form a triangle?

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Well, which combinations will give you a triangle? (They're easy to enumerate in this case.) How many total combinations are there? – cardinal Dec 18 '11 at 3:04
2,3,4,7,8,11: where's the seventh stick? – Robert Israel Dec 18 '11 at 3:25
+1 for the title! – Dilip Sarwate Dec 18 '11 at 3:25

Count the number of three-element subsets of $\{2,3,4,7,8\}$ which are the sides of a triangle. To do this, you can list them by the length of the longest side, to keep track of what you're doing. Keep in mind that $a,b,c$ (with $a\leq b \leq c$) form the sides of a triangle if and only if $a+b\geq c$.

Now the probability you're looking for is the ratio of the number of such subsets, to the total number of three-element subsets.

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It depends on how you define triangle - you might restrict yourself to the case $a+b>c$ – Thomas Andrews Dec 18 '11 at 3:25

"A straight line is the shortest distance between two points."

So if it's 2 cm from A to B and 3 cm from B to C, for a total of 5 cm, then directly from A to C---a straight line---is no longer than 5cm. That means it cannot be 11 cm. Nor 8, nor 7. That's how you tell if they form a triangle.

There are 20 ways to choose 3 out of 6. Look at each of the 20, and see whether they can form a triangle. Then count them.

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