 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Half a Circle, Half a Sphere (Posted on 2009-10-28) Three points on the circumference of a circle are chosen at random. What is the probability the points are all on the same semicircle?

Four points on the surface of a sphere are chosen at random. What is the probability the points are all on the same hemisphere?

 See The Solution Submitted by Brian Smith Rating: 3.0000 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) First part | Comment 2 of 4 | For simplicity, I'll use the unit circle (center at the origin with radius 1) with points A, B, and C as the random points on this circle.  Then fix point A at 0 degrees (point (1,0)) with point A' at 180 degrees (point (-1,0)).  Then point B is free to lie anywhere between -180 and +180 degrees uniformly.  At one extreme, suppose point B coincides with point A -- then point C must lie between -90 and 90 degrees for all points to lie on the same semicircle, so that probability is 1/2.  At the other extreme, suppose point B coincides with point A' -- then point C can lie anywhere on the circle and still have all points on the same semicircle, so that probability is 1.  Therefore the probability of all three points lying on the same semicircle is the average of these two extremes, so that the probability p = 3/4.
 Posted by Jim Keneipp on 2009-10-28 11:32:25 Please log in:

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