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anonymous
 4 years ago
n points are placed at random on the circumference of a circle, what is the probability that they all lie within a common semicircle?
anonymous
 4 years ago
n points are placed at random on the circumference of a circle, what is the probability that they all lie within a common semicircle?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have googled this question and there are many explanations, but i have to say i don't even really understand what it is asking

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0think of placing points at random, now what is the probability that it is possible to cut the circle in half leaving all the points on one side, for example it is possible here: dw:1333717099690:dwbut not here:dw:1333717241684:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0if n=2 P=1 if n>3 P<1 ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ill post my thoughts so far.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0this question reminds me of Buffon's needle problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P(2) is also easy, but we can learn from it: 1 point, A is already there, so placing another point the only place which is questionable is 180 degrees from A , i dont know whether it counts but it doesnt matter as the probability of that position exactly is 0 , therefore P(2) =1

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0the first two points are in the same semi circle always , the third is either inbetween the or outbetween them, equal chance/?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i think so! maybe.. dw:1333724818545:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0placing C on the circle somewhere.. where is allowed?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe its not 1/2 ...

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.01/2 in the worst case scenario, ie if the first two points are opposite

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1333725166177:dw i think its \[\frac{l}{(circumference)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0those lines from A and B are diameters

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0this is a good question . and i can see you are getting closer and closer, but for now i must go (you might want to check out the Buffon's needle problem for some hints

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2I agree, good question This is perhaps a good question for the metamath section

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to anyone viewing i think (guessing) maybe i should find l in terms of theta , then use an integral to find P(theta from A diameter)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whats the meta math section?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2click the "mathematics" blue bar you will see it is a category

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2hard and irregular questions are found and solved there

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2but it can take a long time to get a response in that section, so take your pick keep "bumping" it here, or post in metamath and wait...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2in the meantime let me call on some who may be able to solve this: @across @JamesJ @Zarkon @Mr.Math @FoolForMath interesting probability problem (none are online right now it seems)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2I'm not sure if you can post in both sections with the new "bump" system, but feel free to try :)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2oh, mr.math is here after all maybe he has some nice thoughts on this

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0I found this http://mathproblems.info/images/prob1.pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I liked this answer: http://math.stackexchange.com/a/18371/2109

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2I can almost understand the MSE one, but I can't read Mr.Math's answer... when will my brain grow up like that?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2I can see how they're kinda the same...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i understand FFMs but i am struggling with mr.maths , i understand what's going on in general, im just not following all of the maths there. i think i need more experience with continuous probability and expected values..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gonna close it up now unless anyone has more to add, thanks guys for all your help :)
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