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Jim, not sure if I'm missing something, but wouldn't you need four points on a sphere to create a tetrahedron? Or is it taken as read that the first point of four can effectively be treated as being fixed? Cheers, Ian p.s. As for the probability, nice question, not sure! I have an initial hunch that the second, third and fourth points would need to be on the surface of the hemisphere directly opposite the first point to encompass the origin, so on a simple random probability, wouldn't the answer just be 1/8? Not sure how the normality you mention is relevant. You may leave the list at any time by sending the command SIGNOFF allstat to [log in to unmask], leaving the subject line blank.