ganeshie8
  • ganeshie8
A triangle is formed by picking \(3\) points at random from the vertices of a regular \(2n+1\)-gon. Find the probability for it to be an equilateral triangle.
Mathematics
jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
for first few values of \(n\) im getting below probabilities : n = 1, probability = 1 n = 2, probability = 0 n = 3, probability = 0
thomas5267
  • thomas5267
My guess would be for 3n-gon, \(P=\dfrac{n}{\binom{n}{3}}\) and P=0 for all other regular polygon.
ganeshie8
  • ganeshie8
Thats right! could you share how you arrived at that

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ganeshie8
  • ganeshie8
pretty sure you meant \(P=\dfrac{n}{\binom{\color{red}{3}n}{3}}\)
thomas5267
  • thomas5267
I am absolutely confused now lol. I thought the answer would be \(\dfrac{3n}{\binom{3n}{3}}\). For all regular polygon of not having 3n vertices, rotation of 120° would not constitute a valid symmetry. So that restricts us to 3n-gon. For 3n-gon, there are \(\binom{3n}{3}\) choices of 3 vertices. Now I have to check my numerator!
thomas5267
  • thomas5267
Ah yes, \(\dfrac{n}{\binom{3n}{3}}\) is indeed the correct answer as there are n equilateral triangle in 3n-gon.
ganeshie8
  • ganeshie8
Looks nice, there are just some repetitions to be accounted for..

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