Pick two points at random on a circle and draw the chord connecting them.
Pick two more points and connect them with a second chord.
What is the probability that these chords intersect?
(In reply to
solution, maybe by Stephanie)
I think, looking at your solution, it's equivalent to one I had in mind, that, at least to me, gives me greater assurance that all the possibilites that are taken have equal probability:
Call the chords AB and CD.
Choose point A. Then three more points will be chosen randomly and independently. Seen clockwise from A, the order of these chosen points is equally likely (since there is nothing making one order more likely than another) to be BCD, BDC, CBD, CDB, DBC or DCB. Only if B is between the other two do the chords cross, and that is 2/6 = 1/3 probability.
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Posted by Charlie
on 2009-05-11 22:56:01 |
re: solution, maybe
