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CAF123
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Homework Statement
Suppose that n points are independently chosen at random on the circumference of a circle and we want the probability that they all lie in some semicircle.
Let ##P_1...P_n## denote the n points. Let A denote the event that all the points are contained in some semicircle and let ##A_i## be the event that all the points lie in the semicircle beginning at the point ##P_i## and going clockwise for 180 degrees, ## i \in {1,...n}## Find P(A).
The Attempt at a Solution
I know $$ P(A) = P(\cup_{i=1}^{n} A_i) = \sum_{i=1}^{n} P(A_i) $$ since the ##A_i## are mutually exclusive. I then said each ##P_i## is a uniformly distributed point with density $$f(x) = 1/(2\pi), x \in (0,2\pi). $$
So $$P(0≤x≤\pi) = \int_{0}^{\pi} 1/(2\pi) dx = 1/2. $$ So this is the probability of one point being in a semicircle. (this result being obvious) So for the first point, I have $$P(A_1) = 1/2 (1/2)^{n-1}. $$ Similarly, $$P(A_2) = (1-1/2)(1/2)(1/2)^{n-2} $$ and $$P(A_3) = (1-1/2)(1-1/2)(1/2)(1/2)^{n-3} ...$$
Bringing this together, I get $$P(A) = P(A_1) +P(A_2) +...P(A_n) = n(1/2)^n, $$ after simplification. But the answer says ##n(1/2)^{n-1}?## Any advice? Many thanks
EDIT: I thought it through a little more. The probability that the first point starts the semicircle is 1/n. The probability of the remaining n-1 points lying in the semicircle is ##(1/2)^{n-1}## Would ##P(A_1) = 1/n (1/2)^{n-1}?##
EDIT2: Sorry, misread the Q. ##P(A_1)## is the probability that all the points lie in a semicircle beginning at point 1. This is a given, but I previously assumed they wanted to know the probability that ##P_1## was the first point. Hence, fixing point 1, the probability that all the remaining points lie in the semicircle going clockwise is 1/2 for each point (either it is in a semicircle going clockwise or anticlockwise) So ##P(A_i) = (1/2)^{n-1} => P(A) = n (1/2)^{n-1}.## Is this now a reasonable argument?
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