1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability that n points lie on one side of a circle

  1. Nov 19, 2012 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    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).

    3. 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?
     
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks like you got there. The argument is sound.
     
  4. Nov 19, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you use the results of problem 18, page 41 of Feller, "An Introduction to Probability Theory and its Applications", Vol. II (Wiley, 1971), the answer will come out as
    [tex] P_{Feller} = \frac{n+1}{2^n}, \: n \geq 3 .[/tex] For n = 3 this gives a probability of PFeller = 1/2, while the answer you wrote gives 3/4. (Note: the n in Feller is the same as the n in your problem!)

    RGV
     
  5. Nov 19, 2012 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    For n=3, I believe 3/4 is correct. We can fix one of the points as an origin and measure the locations of the other two from there as being in the range [0, 1], uniformly distributed. We can represent the pair graphically as (x, y), the possibilities forming a unit square.
    If x<0.5 and y<0.5, the condition is met.
    If 0.5<x and 0.5<y, the condition is met.
    If |x-y| > 0.5 the condition is met.
    These are non-overlapping areas adding up to 3/4.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability that n points lie on one side of a circle
  1. N points on a circle (Replies: 1)

  2. Points on a circle (Replies: 3)

Loading...