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Problem of the Week #55 - April 15th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: A point is uniformly distributed within the disk of radius 1. That is, its density is\[f(x,y)=C,\qquad 0\leq x^2+y^2\leq 1\]
Find the probability that its distance from the origin is less than $x$, $0\leq x\leq 1$.

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Note​: $f(x,y)$ is a density function if $\displaystyle\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\,dy\,dx = 1$.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by Ackbach, alphabeta89, and Bacterius; Sudharaka gets honorable mention since he computed the wrong probability... :-/

Here's Ackbach's solution:

First we normalize the function $f(x,y)$ by discovering $C$. We need that
$$\int_{\mathbb{R}} \int_{\mathbb{R}} f(x,y) \, dx \, dy=1,$$
or, switching to polar coordinates, that
$$\int_{0}^{2 \pi} \int_{0}^{1}C \, r \, dr \, d\theta=1.$$
The integral on the LHS is essentially the area, so we have that
$$C= \frac{1}{ \pi r^{2}}= \frac{1}{ \pi},$$
and therefore that
$$f(x,y)= \frac{1}{ \pi}.$$
Next, to find the probability that the point is within a distance of $x$ to the origin, we must
compute the probability
$$P(r<x)= \frac{( \pi x^{2})/ \pi}{( \pi)/ \pi}= x^{2}.$$

Note that I could probably have solved this problem without normalizing, but I felt like doing it.

And here's Bacterius' solution:

[JUSTIFY]Let $\text{X} : (\theta_\text{X}, r_\text{X})$ be a uniformly selected point on the unit circle, in polar coordinates. What is the probability that $r_\text{X} < x$ for some $0 \leq x \leq 1$?

Let us divide the unit circle in concentric rings with inner radius $r$ and thickness $\text{d} r$. What is the total area of all rings up to some radius $x$? We have:

$$A_x = \int_0^x 2 \pi r ~ \text{d} r = \pi x^2$$
This area $A_x$ is the area of the circle of radius $x$ concentric to the unit circle, enclosing all points which have distance less than $x$ to the origin. And the total area of the unit circle is equal to $A = \pi$, so the probability of $\text{X}$ falling inside the area $A_x$ is equal to (this is really a cumulative distribution function):

$$P(r_\text{X} < x) = \frac{A_x}{A} = \frac{\pi x^2}{\pi} = x^2$$
We conclude, that a uniformly selected point on the unit circle has probability $x^2$ to have distance less than $x$ to the origin.[/JUSTIFY]
 
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