 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Same side story (Posted on 2017-03-16) Suppose n (n>1) points are placed at random on a circumference of a circle.
If P(n) denotes the probability that all n points lie on the same side of some diameter - find P(2), P(3) and P(4).

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Corrected solution | Comment 3 of 5 | Sorry, my previously proposed solution is wrong, and so is also Jer's calculation of P(4). Here is my new one :
Like Jer,we define the points in polar coordinates so that the first one lies on the x axis with the angular value of  0 . The second point will have the value of 0<phi<180, and clearly P(2)=1.
In order for the 3rd point to lie on the same half circle, it is easily seen that the 3rd point must be either between 0 to 180, or between (180+phi) and 360, which allows it to lie along 180+(180-phi)=(360-phi) degrees.
The probability of the 3rd point to answer the problem's condition is therefore (360-phi)/360.
phi itself is a random number being equally likely between 0 and 180. Its expectation is therefore E=180/2=90. Substituting E =90 instead of phi,we get the probability P(3)= (360-90)/360=0.75.
As for the 4th point, after the first 3 ones have been placed, we choose the 2 extreme ones, define now phi as the angular difference between them, and repeat exactly the same process we did for the 3rd point. The outcome will therefore agian be that the probability of the 4th point to answer the problem is 0.75.
But now we must multiply this probability by the result fot the 3  first points, and we get P(4)=0.75*P(3)= 9/16 .

 Posted by Dan Rosen on 2017-03-28 23:24:18 Please log in:

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