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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