Inside a circle, radius 1, 10 points are chosen. $X$- random variable that represents the distance from the rim of the circle to the closest point. Find the distribution of $X$. My thoughts: For $P\{X < t\}$ I need to find the probability that at least one point out of the ten occupies the ring whose inner radius is $1-t$ and outer is $t$. I was thinking of finding this my finding the probability that no points of the ten occupy this ring. $$1-P\{A\}=1-\frac{(1-t)^{20}\pi^{10}}{\pi^{10}}$$
 A: Let $(X,Y)$ be a uniformly distribution within the unit disk point. Then
$$
    f_{X,Y}\left(x,y\right) = \frac{1}{\pi} \cdot [ 0 <x^2+y^2 <1]
$$
where $\left[ \,\cdot\,\right]$ denotes Iverson bracket. The distribution of distance of this point to the origin $R = \sqrt{X^2+Y^2}$ is (e.g. here):
$$
    f_R\left(r\right) = 2 r \cdot [ 0<r<1 ]
$$ 
Hence the distribution of the distance from the point to the rim of the circle, $R^c = 1 - R$ is 
$$
    f_{R^c}\left(\rho\right) = 2 \left(1-\rho\right) \cdot [ 0 < \rho < 1 ]
$$
Clearly $F_{R^c} \left(\rho\right) = \Pr\left(R^c \leqslant \rho\right) = 1 - \left(1-\rho\right)^2$ for $0<\rho<1$.
The smallest distance $D$ to the rim from $n$ independent and uniformly distributed within the disk points is the smallest order statistics.
$$ \begin{eqnarray}
      F^c_D\left(d\right) &\stackrel{0<d<1}{=}& \Pr\left( \min\left(R_1^c, \ldots, R_n^c\right) > d \right) \\ &=& \Pr\left( R_1^c > d, \ldots, R_n^c > d \right) = \prod_{k=1}^n \Pr\left(R_k^c > d\right) \\
  &=& \Pr\left(R^c > d\right)^n = \left(1-d\right)^{2n}
\end{eqnarray}
$$
Hence the density of $D$:
$$
    f_D\left(d\right) = 2 n \left(1-d\right)^{2n-1} \cdot \left[ 0 < d < 1 \right]
$$
A: Let $Y$ be a random variable representing the maximum distance of any point from the origin. That is, $Y = 1-X$. That shift makes it way easier to think about things. 
$P(Y<r)$ is just how likely it is to choose all 10 points within a circle of radius $r$, so $P(Y<r) = (r^2)^{10}$. To find:
$$\begin{split}P(X < r) &= P(Y > 1 - r)\\
&= 1 - P(Y < 1-r) \\
&= 1 - (1-r)^{20}
\end{split}$$
So the PDF of $X$ is $20(1-r)^{19}$

Another way of thinking about this problem: $X$ is between $r$ and $r+dr$ with probability $f_X(r)dr$. That would be one point to be between $(1-r)$ and $(1-r)+dr$ and the other 9 inside of $(1-r)$. Which is: 
$$\begin{split}
f_X(r)\cdot dr &= \binom{10}1\cdot((1-r+dr)^2-(1-r)^2)\cdot((1-r)^2)^9 \\
&= 10 \cdot (2(1-r)dr + dr^2)(1-r)^{18} \\
&= 20dr(1-r)^{19} + O(dr^2)
\end{split}
$$
We can drop the $dr^2$ term since it's negligible, so we end up with $f_X(r) = 20(1-r)^{19}$, as before.
