# Density function problem?

• Jul 20th 2011, 02:13 AM
downthesun01
Density function problem?
I'm not even sure if that's the proper thing to call it, but here's the question.

A point is uniformly distributed within the disk of radius 1. That is, its density is
f(x,y)=C, 0≤x^2+y^2≤1
Find the probability that its distance from the origin is less than x, 0≤x≤1.

I'm not sure of what I'm supposed to do here. What are my steps?
• Jul 20th 2011, 03:08 AM
chisigma
Re: Density function problem?
Quote:

Originally Posted by downthesun01
I'm not even sure if that's the proper thing to call it, but here's the question.

A point is uniformly distributed within the disk of radius 1. That is, its density is
f(x,y)=C, 0≤x^2+y^2≤1
Find the probability that its distance from the origin is less than x, 0≤x≤1.

I'm not sure of what I'm supposed to do here. What are my steps?

The requested probability is the ratio between the area of a circle af radious $\displaystyle \rho$ and the circle of radious 1...

$\displaystyle P \{d < \rho\} = \frac{\pi\ \rho^{2}}{\pi}= \rho^{2}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jul 20th 2011, 09:01 PM
downthesun01
Re: Density function problem?
I still don't understand the question. I'm trying to picture it right now.

There's a point that's uniformly distributed within a circle that has a radius of 1, and I'm supposed to find the probability that the point's distance from the origin of the circle is less than x.

I don't understand where this second circle with a radius of rho comes from.

I can solve for C. That seems rather straightforward.

$\displaystyle 1=\int \int_R Cdxdy=c\pi r^2=c\pi$
$\displaystyle c=\frac{1}{\pi}$

But from there I'm pretty confused. If someone could offer some insight, I'd appreciate it. Thanks

What does $\displaystyle c=\frac{1}{\pi}$ mean? Is it the probability that the point occupies a certain spot in the circle?

And what is x supposed to be?