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

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Its 3/4, in geometrical probability we neglect boundary points hence right angled triangle case neglected.
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Exam patterns and general useful info and tips about the exam.
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Exam patterns and general useful info and tips about the exam.
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Exam patterns and general useful info and tips about the exam.
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Its 3/4, in geometrical probability we neglect boundary points hence right angled triangle case neglected.
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joining one would do good because at coaching you can get the idea of exam patterns and general useful info and tips about the exam.
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joining one would do good because at coaching you can get the idea of exam patterns and general useful info and tips about the exam.
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Let f_n(x) the density probability of the span of n points, that is the smallest sector which contains them. For n = 2, f_2(x) = 1/ pi for x in and 0 beyond. For n = 3, f_3(x) = ( 2 / 2 pi) *int f_2(u) du + f_2(x) * (x/2pi). This expresses that, to have a span x for three points, you start...
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Let f_n(x) the density probability of the span of n points, that is the smallest sector which contains them. For n = 2, f_2(x) = 1/ pi for x in [0,pi] and 0 beyond. For n = 3, f_3(x) = ( 2 / 2 pi) *int [0,x] f_2(u) du + f_2(x) * (x/2pi). This expresses that, to have a span x for three points, you start from 2 points: either their span is less than x, and you have 2 positions where to put the third point, or their span is x, and you have the sector between the 2 first points available. This formula remains valid if one replaces f_2 and f_3 by f_n and f_(n+1) For x< pi, f_n(x) is of the form C_n x^(n-2) where C_n can be computed by induction. In the end your probability P_n = integral [0,pi] f_n(x) dx = n / 2^(n-1). read less
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Hi Kunal, May I know, How did you arrive to the answer? 0.5 seems to be matching my answer
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Its 3/4, in geometrical probability we neglect boundary points hence right angled triangle case neglected.
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