1

Probability that a uniformly selected point in the interior of sphere of a given arbitrary radius (for the purposes of this, we will say 2022) is closer to the surface of the sphere than the center.

I was imagining taking a slice of the sphere and calculating the probabilities from there, but I can not seem to follow any clear logic.

Christian E. Ramirez
  • 3,377
Kevin Perez
  • 29
  • If the point is closer to the centre than it is to the surface, then it lies within a sphere half the radius of the original. What is the ratio of volumes between the two spheres? – Théophile Jan 06 '23 at 02:10
  • @Théophile I am a little bit confused about what you mean by between the two spheres, there is only one sphere and the answer is 7/8, but I can not wrap my head around how this is the case. – Kevin Perez Jan 06 '23 at 02:12
  • By the "two spheres" I meant the given one in the problem and the other that I mentioned, "a sphere half the radius of the original." – Théophile Jan 06 '23 at 02:14
  • In $d$ dimensions, the ball with radius $r$ has volume $r^dV(B(0,1))$, where the second term is the volume of the unit ball. Can you show this? – Andrew Jan 06 '23 at 02:15
  • Well I would suppose they would be the cube of the multiple, i.e 8. – Kevin Perez Jan 06 '23 at 02:15
  • That's right; so the smaller sphere is $\left(\frac12\right)^3 = \frac18$ the size of the larger. That's the probability of being close to the centre, so the probability of being close to the surface is $1 - \frac18$. – Théophile Jan 06 '23 at 02:16
  • Wow, This makes perfect sense. I did not see such as clear and simple answer before. – Kevin Perez Jan 06 '23 at 02:19

0 Answers0