| Hard | Probability |
What is the expected distance of any point on Earth and the north pole? Take Earth radius 1.
Clarification: Shortest distance cuts through the sphere, instead of lying on surface.
Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?
Clarification: Shortest distance cuts through the sphere, instead of lying on surface.
Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?
Hint
Imagine a ring of some thickness, whose distance from N is constant from all points on that ring. Get the average of all such rings.
Answer
4/3
Solution
2*sin(x/2)is the distance of north pole from a point on the ring at angle x from the z axis. So, integrate from 0 to pi, (2*pi*sin(x)*2*sin(x/2) dx) and divide by the total area which is 4*pi.
Answer:4/3
Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole).
Area of ring = 2*pi*dy
Probability of choosing point on this ring = dy/2
Distance of N & a point on ring = sqrt(2y)
Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3
And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question.
Answer:4/3
Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole).
Area of ring = 2*pi*dy
Probability of choosing point on this ring = dy/2
Distance of N & a point on ring = sqrt(2y)
Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3
And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question.
Source: Quant Interview
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