Any ideas on how to solve it?


The hemisphere is not predefined. It can be any hemisphere.


closed as unclear what you're asking by Did, John B, colormegone, Chill2Macht, Leucippus Jun 13 '16 at 0:01

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  • $\begingroup$ Is the hemisphere chosen ahead of time? $\endgroup$ – Semiclassical Jun 6 '16 at 17:36
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    $\begingroup$ It depends a bit what you do between the different shots. Typically in interviews, you are supposed to check with the interviewer and ask him/her to clarify. If you don't move the gun between the shots then I would guess the probability is 100%. $\endgroup$ – Fabian Jun 6 '16 at 17:37
  • $\begingroup$ I have edited the question @Semiclassical $\endgroup$ – user2331 Jun 6 '16 at 17:38
  • $\begingroup$ Thanks. I'm not sure I'm visualizing it right, but if the hemisphere isn't chosen ahead of time (i.e. 'what is the probability that all three end up in 'some hemisphere') then I can hardly see how it wouldn't always be true. $\endgroup$ – Semiclassical Jun 6 '16 at 17:39
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    $\begingroup$ There is a difference between saying "Three bullets are shot in a sphere" and "Three bullets are shot into a sphere". $\endgroup$ – DanielV Jun 7 '16 at 8:17

1 because three points define a plane (P) that does not in general contain the origin (we set apart the cases where this plane contains the origin $0$, which is of probability 0). Then it suffices to consider the plane parallel to (P) passing through the origin : it divides the sphere into two hemispheres, one of which contains the three points.


It probably also depends on the way you choose (shoot at) those three points. Similarly to Bertrand's paradox.

For example (referring back to @JeanMarie's answer), if you force the third bullet to make the plane of the three bullets go through the centre of the sphere, that certainly is a valid probability distribution. One might argue that's not random enough, but in any way that gives you a probability of zero.

  • $\begingroup$ In order to identify this as having "probability of zero" a probability distribution needs to be defined. Forcing something (such as having the bullets travel in a common plane) that you say has probability zero then produces a conditional probability distribution (which must have total probability one, given the restricted circumstance). $\endgroup$ – hardmath Jun 7 '16 at 8:26
  • $\begingroup$ I'm not sure if I get your point. By forcing I meant I could come up with a probability distribution which would produce such a result every time. E.g. putting a delta-function at the origin. $\endgroup$ – zahypeti Jun 7 '16 at 8:38
  • $\begingroup$ Actually, I just realised that we also have to agree on how to count those on the mid-plane (as in, is it a solid hemisphere with the boundaries or a hollow one) $\endgroup$ – zahypeti Jun 7 '16 at 8:40
  • $\begingroup$ Perhaps you are saying we should accept JeanMarie's Answer, given the limited information in the "interview" Question, with the rationale that either three bullets colinear with the center counts as being "in the same hemisphere" or amounts to an event of probability zero. $\endgroup$ – hardmath Jun 7 '16 at 12:18

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