In two dimensions, the probability that all n molecules lie in some semicircle is 2n/2^n.
That doesn't look right. n=2 -> P=1
Looks right to me. Given a point inside a circle other than the center, all points inside the circle are in some open semicircle with the original point except the center and radius opposite the original point. (All points share some closed semicircle with the original point.) Charles Greathouse Analyst/Programmer Case Western Reserve University On Tue, Sep 17, 2013 at 3:19 PM, meekerdb <meekerdb@verizon.net> wrote:
On 9/17/2013 12:02 PM, Eugene Salamin wrote:
A spherical vessel contains n molecules of gas. What is the probability that all the molecules can be found in one hemisphere? For a given hemisphere, it is 1/2^n. For 6 hemispheres arranged like the faces of a cube, inclusion-exclusion gives 6/2^n - 12/4^n + 8/8^n. But what is the probability if any hemisphere is allowed? I'm stuck on this problem.
I don't understand the arrangement. Are the hemispheres stuck onto the faces of a cube?
In two dimensions, the probability that all n molecules lie in some semicircle is 2n/2^n.
That doesn't look right. n=2 -> P=1
A related question is: Given n vectors x[1], ... , x[n], how do we test if they all lie in some half-space, i.e. does there exist a vector u such that u.x[i] > 0 for all i? I'm stuck on this as well.
If the boundary of the half-space is a plane, can't you just take the inner product with a vector normal to that plane?
Brent
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