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What is the probability that three points, randomly put on a circle will fall in the same semicircle?

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n/(2^(n -1)) = 3/2^2 = 3/4
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M.Sc Statistics

1/8
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M.Sc Statistics

there are two semi cicles, probability that a random a point falls in any one of the semi circles is 1/2.Here we are doing three random draws independent of oneanother so 1/2*1/2*1/2=1/8.
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Suppose that point i has angle 0 (angle is arbitrary in this problem) -- essentially this is the event that point i is the "first" or "leading" point in the semicircle. Then we want the event that all of the points are in the same semicircle -- i.e., that the remaining points end up all in the upper...
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Suppose that point i has angle 0 (angle is arbitrary in this problem) -- essentially this is the event that point i is the "first" or "leading" point in the semicircle. Then we want the event that all of the points are in the same semicircle -- i.e., that the remaining points end up all in the upper half plane. That's a coin-flip for each remaining point, so you end up with 1/2n?1. There's n points, and the event that any point i is the "leading" point is disjoint from the event that any other point j is, so the final probability is n/2n?1 read less
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3/4 for generic n point it is n/2^(n-1)
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Trainer

I think, the answer is 1/8
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M.A. In Geography, Appeared in M. Phil

final probability is n/2^n?1
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