Finding Geometric Probability? Four points are chosen at random from the interior of a circle. What is the probability that the sum of each of their distances from the center is greater than half the circumference of the circle?
I don't even know how to begin this problem. Any hints would be greatly appreciated.
 A: Guide:


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*Find the distribution of $R=\sqrt{X^2+Y^2}$ where $(X,Y)$ is chosen uniformly on the disk $\{(x,y)\in\mathbb R^2\mid x^2+y^2<1\}$. It is handsome to go for the CDF first. A PDF can be found by differentiating.

*Find $\Pr(R_1+R_2+R_3+R_4>\pi)$ where the $R_i$ are independent and have the same distribution as $R$.
A: Assume the disk have radius $1$.
If you have $r \sim Uni[0, 1]$, i.e. random radius and $\theta \sim Uni[0, 2\pi]$, i.e. random angle, then $X = \sqrt{r} \cos \theta$ and $Y = \sqrt{r} \sin \theta$ have a joint pdf on the disk $\frac{1}{\pi}$, i.e. describes a point picked uniformly at random on the disk. 
And $R = \sqrt{X^2 + Y^2}$.
A: Denote by $R$ the absolute value of a uniformly distributed random point $Z$ in the unit disc. The cdf and the pdf  of $R$ are then given by
$$F_1(s)=\left\{\eqalign{0\quad&(s<0) \cr s^2\quad&(0\leq s\leq1)\cr 1\quad&(s\geq1)\cr}\right.\ ,\qquad
f_R(s)=\left\{\eqalign{0\quad&(s<0) \cr 2s\quad&(0< s< 1)\cr 0\quad&(s>1)\cr}\right.\ .\tag{1}$$
The index ${}_1$ refers to the first random point. Using $(1)$ one computes the cdf $F_2(s)$ of $R_1+R_2$ for two random points, and then $F_3(s)$, $F_4(s)$ as follows:
$$F_2(s)=\int_0^1 f_R(t) F_1(s-t)\>dt,\quad\ldots, \quad F_4(s)=\int_0^1 f_R(t)F_3(s-t)\>dt\ .$$
Mathematica obtained the following expression for $F_4(s)$ in the interval $3<s<4$:
$${31192 - 12288 s - 17920 s^2 + 12544 s^3 - 1680 s^4 - 448 s^5 + 
 112 s^6 - s^8\over2520}\ .$$
The probability $p$ that $R_1+R_2+R_3+R_4\geq\pi$ then comes to
$$p=1-F_4(\pi)\approx0.1625\ .$$
