You believe that there's a fly somewhere less then 6 feet away from you.

Question

You believe that there is a fly somewhere less than $6$ feet away from you. If you believe that he is located Uniformly in a circle of radius $6$ feet away from you, what is the probability that he is more than $2$ feet away from you?

My work: This is clearly a continuous uniform distribution question, and I know that we need to account for area here for the density function since the fly is within a circular radius. So I believe we use the joint pdf $f(x,y) = \cfrac{1}{36\pi}$ to help us calculate the probability the fly is more than $2$ feet away from me since if I integrate over the area of the circle I get $1$. But I don't know where to go from here and how I can use this info to help me calculate the probability. Also what would the bounds of $f(x,y) = \cfrac{1}{36\pi}$ be? I know that $(x,y)$ would be whatever's in the circle, but what would that be?

Any help with finding the bounds or probability would be greatly appreciated! Thank You!

Answer 1

First do a drawing

Your probability is simply the "favourable area" (purple area) divided by the "total area" (big circle area). Integration is not needed because the distribution is Uniform. Anyway if you want to integrate you will find the same result. Try it!

Simply do the ratio of the 2 areas to get the probability. It is easier to calculate the probability that the fly is LESS than 2 feet from you.... thus you requested probability is the complement

$$P[X>2]=1-\frac{\pi 2^2}{\pi 6^2}=1-\frac{4}{36}=\frac{8}{9}$$

Answer 2

Just for fun, you could do an integral in polar coordinates, where $dA=rdrd\theta$. That is how you would do it using the "uniform density".

$\displaystyle \int_0^{2\pi}\int_2^6 \frac r {36\pi} dr d\theta=\frac 1 {36\pi}\int_0^{2\pi}16d\theta=\frac{32\pi}{36\pi}=\frac 8 9$