Each plane is denoted by Ax + By + Cz + D = 0, where A, B, C, D are real numbers (specific to that plane). --
I suspect the answer is P=0, since the area outside any bounded region must exceed that inside. But how to prove that?
Nitin P. answered • 06/28/20
Machine Learning Engineer - UC Berkeley CS+Math Grad
A plane divides 3-D space into 2 equal halves. So, a point has a 50% chance of being on either side of the plane. Therefore, the probability of being in a region bounded by 3 planes is (1/2)3 = 1/8. Note that your answer of P = 0 is actually a conditional probability, conditioning on the planes not being parallel and forming a bounded region.
Rabie R. answered • 06/28/20
Experienced, Patient and Knowledgeable Math Tutor
This question is ill-posed. What is the underlying distribution of the points in the 3-D space?
Note that it cannot be a uniform distribution because 3-D space is unbounded.
Rabie R.
Your answer makes sense if the question is reformulated. However, as the question is formulated now, there are issues with it: 1) you need a distribution. The points are fixed, but are all points equally likely outcomes when you're picking a point? 2) The question asks about being in a region "bounded" by 3 planes. The question doesn't specify a distribution, so assuming all points are equally likely to be picked, you can never get a 1/8 probability of being inside a finite volume out of a domain with an infinite volume.06/28/20
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Cindy M.
5.0 (1,320)
Maya K.
5 (105)
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5 (800)
Nitin P.
We don't care about the distribution of the points, because they're already fixed in space and deterministic. We care about what happens when we randomly place the planes.06/28/20