IIT JEE , AIEEE Forum - Mathematics , Algebra

akhil_o

A point is chosen from a sphere of radius 5(on or inside)

What is the probability that the point lies on the surface of the sphere?

man111

I think the inswer should be 4pi*r^2/4/3pi*r^3=3/r=3/5

hsbhatt

The previous answer is right. First imagine the points as a mesh spread over the surface. Thus we have a surface density call it

and it can be made infinitely large. Thus the number of points at a radius r =

*4

r². The total number of points can be got as ₀

^R

*4

r²dr = 4

R³/3.

Hence Prob (point at radius r) =

*4

r²/(4

R³/3) = 3r²/R³.

Here r = R = 5

Hence Prob (point at surface) = 3/5

T_T

zero......volume of surface is zero..but tat of inside portion is finite

nadeemoidu

T_T is right.
The answer is 0 and can be got by using pure common sense. It is quite obvious that a point chosen at random from a sphere cannot be exactly on the surface.

@hsbhatt
There is no density involved here.
the probability = favourable volume / total volume.

Favourable volume =area x height= 4

r² dr
But since dr is infinitely small , this quantity will be 0. dr can be assumed to be non-zero only when we are using integration.

Total volume = 4

R³/3.
Since this is finite , the final answer will be 0.

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