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ganeshie8
 one year ago
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.
ganeshie8
 one year ago
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.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for first few values of \(n\) im getting below probabilities : n = 1, probability = 1 n = 2, probability = 0 n = 3, probability = 0

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1My guess would be for 3ngon, \(P=\dfrac{n}{\binom{n}{3}}\) and P=0 for all other regular polygon.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Thats right! could you share how you arrived at that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1pretty sure you meant \(P=\dfrac{n}{\binom{\color{red}{3}n}{3}}\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I 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 3ngon. For 3ngon, there are \(\binom{3n}{3}\) choices of 3 vertices. Now I have to check my numerator!

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Ah yes, \(\dfrac{n}{\binom{3n}{3}}\) is indeed the correct answer as there are n equilateral triangle in 3ngon.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Looks nice, there are just some repetitions to be accounted for..
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