# Probability Question

1. Oct 26, 2005

### Edwin

I came up with the following question, and wanted to see if I have the right solution.

Suppose that the probability that a particle is on or within a sphere is
100%.

I figured that the probability that a particle is in a sphere is the volume of the a sphere of radius r, based on what we covered in our calculus class. The probability that a particle is on the surface of a sphere of radius r is simply the surface area of the sphere of radius r.

I presume that the probability that a particle is in or on a sphere(of radius r) is the surface area of the sphere plus the volume of the sphere. Thus if the probability of a particle is in a sphere or on a sphere is 100%, then the probability function for the particles position is simply

(Area of sphere) + (Volume of Sphere) = 1

4pi(r^2) + (4/3)(pi)(r^3) - 1 = 0 which has one solution of r is approximately equal to .27019.

In other words, since the volume and surface area of a sphere is a function of one variable r, then there turns out to be only one solution of r. The radius of the sphere is is, r is approximately 0.27019.

What do you think? Does this sound correct?

Inquisitively,

Edwin

Last edited: Oct 26, 2005
2. Oct 26, 2005

### Edwin

I thought that if this were correct, than it would be a universal function.

Consider a single particle that enters a basic non-rotating black hole with a spherical event horizon. Any particle that arrives at the event horizon would have a 100 percent probability of being in the black hole, or on the event horizon of the black hole. If the black-hole had only one particle, it seems this would imply the radius of the black hole would have to be 0.27019. I'm not sure what that means, or what the units of measurement would be on the radius.

Incidently, if the probability that the particle is in the volume of the black hole is 100%, then the probability that it is on the event horizon of the black hole is 0%, which still satisfies the probability function specified in the previous post.

(Area of sphere) + (Volume of Sphere) = 1, or rather,

(Surface area of sphere #1) + (Volume of Sphere #2) = 1

Where the probability of the particle being on the area of some sphere S1 is the surface area of the sphere S1 of radius r1, and the probability of the particle being in the volume of sphere S2 of radius r2 is the volume of sphere S2. The above case, where r1 = r2 = 0.27019 is just a special case. In all other cases, where the probability that a particle is in a sphere the radius r, where r does not equal 0.27019, then there must exist two spheres S1 and
S2, of radius r1 and r2, respectively, where either the particle exists in the surface of S1, or it exists in the volume of S2.

Any thoughts?

Inquisitively,

Edwin G. Schasteen

Last edited: Oct 26, 2005
3. Oct 27, 2005

### EnumaElish

I do not quite get how you are adding an area to a volume. I am not sure that you have a sensible measure there. You are assuming that cumulative probability (cdf) is a voluminal measure (otherwise you may not define a probability distribution over a volume). If the sphere is closed then it includes its surface area, hence the surface is not a distinct entity to begin with. If the sphere is not closed, then it doesn't include the surface. But in this case an interior point can be arbitrarily close to the surface, so that the voluminal measure of the surface itself has to be zero.

4. Oct 27, 2005

### mathman

One major flaw in Edwin's approach is that r has units. If you express r in inches or miles or lightyears, you will get different ratios between interior probability and surface probability. In actuality, the volume probability (assuming uniform in volume) already assigns a probability on the surface (=0).

5. Oct 27, 2005

### HallsofIvy

Staff Emeritus
No, it's not correct at all. You started out by saying "Suppose that the probability that a particle is on or within a sphere is 100%. "
It then makes no sense to also say "I figured that the probability that a particle is in a sphere is the volume of the a sphere of radius r".

If you are given that a particle is on or within a particular sphere, then your first statement is true. If you then have another sphere, inside the first (and assume uniform probability) the probability that the particle is within that sphere is the volume of that sphere divided by the volume of the first sphere.
Given that the particle is on or within a particular sphere, the probability that it is within a given subset of the sphere is the volume of the subset divided by the volume of the sphere.
In particular the probability that the particle is on the surface of the sphere is 0: the surface of the sphere has volume 0. (Of course, for continuous probability densities, that does not mean it is impossible.)

6. Oct 27, 2005

### Edwin

Thanks! It didn't look right to me either. I appreciate your input. So what I gather is that I could set up a probability function that is merely the ratio of two volumes.
Assume that the probability that a given particle exists on or within a sphere S1 is 100%, then what I gather you are saying, is that if you take a sphere S2 that is a subset of S1, that is S2 is a smaller sphere inside S1, then the probability that the particle is on or within S2 is simply the ratio of the volume of S2 to the volume of
S1, assuming continuous probability density.
Let Sv1 be the volume of sphere S1, and Sv2 be the volume of S2.
So, in summary, I take it that if the probability that a particle is on or within S1 is 100%, then the probability that the particle is on or within S2 is
Sv2/Sv1, where S2 is a subset contained by S1.
Does this look correct? Is this what you meant?
Inquisitively,
Edwin

Last edited: Oct 27, 2005
7. Oct 27, 2005

### Edwin

That makes perfect sense. The probability of a Probability density function in that has two independant random variables X and Y is the double integral of x and Y with respect to the Area, and is a volume, and is equal to the following:

P = Int(X)dx*Int(Y)dy = Int(Int(x*y)dy)dx, and satisfies the contidions that
f(x,y)>=0, and that the double improper integral from negative infinity to positive infinity of the probability function is equal to 1.

Does this sound accurate?

Inquisitively,

Edwin

8. Oct 27, 2005

### hypermorphism

Yep. It's one way of defining the probability. The probability density function need not be restricted to volume (uniform density). Any function that when integrated over the ball gives 1 and whose integral over any subset within the ball is between 0 and 1 is a potential probability density for points being within regions of the ball.
Consider the 1-ball-with-boundary, [0,1], and the probability density function $$p(x) = \frac{\pi}{2}\sin(\pi x)$$. It characterizes one distribution that lets x have a greater probability of being in a region containing the center.

Last edited: Oct 27, 2005
9. Oct 29, 2005

### Edwin

That makes sense. I seen a similar homework problem, as a matter of fact.
Thanks for the clarification!

Best Regards,

Edwin