Three points are chosen randomly and independently on a circle.
Determine the probability that each of the three pairwise distances between the points is less than the radius of the circle.
The probability that the second point will be within one radius of the first is 1/3, as it has to be within the nearest two arcs out of the six formed by an inscribed regular hexagon.
The third point must lie within the intersection of both such 1/3-circumference arcs about each of the first two points. Its size may range from 1/6 the circumference to 1/3 the circumference, with a uniform probability distribution, so we can use the mean between these extremes: 3/12 = 1/4 is the probability the third lies within one radius of each of the first two given that the first two are within one radius of each other.
The overall probability is therefore 1/3 * 1/4 = 1/12.
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Posted by Charlie
on 2014-08-07 12:22:36 |
solution
