-1
$\begingroup$

What is the probability that a particular set of integer edge lengths selected from an interval $[1,n]$, can form a triangle? That is, let $a,b,c \in \{1,2, \dots n \}$. What is the probability that $a$, $b$ and $c$ are the side lengths to a triangle?

How might this extend to the case where one selects real number edge lengths from the unit interval? Can I look for a cube then exceed the pyramids from it?

$\endgroup$
  • $\begingroup$ I think your question is: What is the probability that 3 lengths selected randomly from a set satisfy the triangle inequality? Are we allowed to select the same length twice? $\endgroup$ – Mason Apr 30 '18 at 1:32
  • $\begingroup$ Yes exactly, and yes we can select the same length twice $\endgroup$ – Jack Apr 30 '18 at 1:48
  • $\begingroup$ Well... Let's start small... $n=1$. Then $a,b,c \in \{1\}$ which means there's an 100% chance that they satisfy the triangle inequality. What about $n=2$? Note that all lengths will satisfy the triangle inequality except when we select exactly one length equal to $2$. That is, $a=b=1, c=2$ fails the triangle inequality. It fails $3$ out of the $8$ cases. $\endgroup$ – Mason Apr 30 '18 at 1:59
  • 2
    $\begingroup$ Duplicate of Probability that a triangle can be formed from a permutation of three edges of random length $\endgroup$ – dxiv Apr 30 '18 at 2:22
0
$\begingroup$

Here is an except from my answer to a similar question:

If A+B > C, you have 100% chance of making a triangle. If A+B=C, the triangle is 2d and does not qualify. If A+B < C, you have three joined line segments that do not meet at two ends. You will have to figure the probabilities of each of these equalities/inequalities and go from there.

$\endgroup$
  • $\begingroup$ If A+B > C, you have 100% chance Not so, try $A=1, B=100, C=10$ for example. $\endgroup$ – dxiv Apr 30 '18 at 2:24
  • $\begingroup$ It should be $A + B > C$ and $|A - B| < C$ $\endgroup$ – Phil H Apr 30 '18 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.