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

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

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