Picking two random points on a disk I try to solve the following:

Pick two arbitrary points $M$ and $N$ independently on a disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2 \leq 1\}$ that is unformily inside. Let $P$ be the distance between those points $P=d(M,N)$. What is the probabilty of $P$ being smaller than the radius of the disk?

Picking a specific point $s$ as the first point $M$ on a disk has probabilty $P(M=s)=0$ since we are looking at the real numbers. My intention, that is probably wrong, is that chosing the second point $N$ such that the distance between them is bigger than the radius of the disk, is as likely as being smaller.

I have absolutely no idea how to solve this.
Let $R$ be the distance between $O$, the origin, and $M$. The probability that $R$ is less than or equal to a value $r$ is
$$P(R\le r) = \begin{cases}
\frac{\pi r^2}{\pi\cdot 1^2} = r^2, &amp; 0\le r\le 1\\
1, &amp;r&gt;1\\
0, &amp;\text{otherwise}
\end{cases}$$
The probability density function of $R$ is
$$f_R(r) = \frac{d}{dr}P(R\le r) = \begin{cases}
2r, &amp; 0&lt; r&lt; 1\\
0, &amp;\text{otherwise}
\end{cases}$$



If $OM = d$, $0\le d\le 1$, then the area of the intersection of the unit circles centred at $O$ and $M$ respectively is

$$\begin{align*}
A(d) &amp;= 4\left(\frac12\cos^{-1}\frac d2\right) - 2\left(\frac12d\sqrt{1^2-\frac {d^2}{2^2}}\right)\\
&amp;= 2\cos^{-1}\frac d2 - \frac 12 d\sqrt{2^2-d^2}\\
\frac{A(d)}{\pi\cdot 1^2}&amp;=\frac{2}{\pi}\cos^{-1}\frac d2 - \frac{1}{2\pi} d\sqrt{2^2-d^2}
\end{align*}$$
The last line is the probability that $N$ is within a unit distance of $M$, as a function of distance $d = OM$.



Combining with the probability density function $f_R$ above, the required probability is
$$p = \int_0^1\left(\frac{2}{\pi}\cos^{-1}\frac r2 - \frac{1}{2\pi} r\sqrt{2^2-r^2}\right)\cdot 2r\ dr \approx 0.58650$$

(WolframAlpha)



$$\begin{align*}
I_1 &amp;= \int_0^1 r\cos^{-1}\frac r2\ dr\\
&amp;= \int_{\pi/2}^{\pi/3} 2\theta\cos\theta\cdot(-2)\sin\theta \ d\theta &amp;&amp; (r = 2\cos \theta)\\
&amp;= -2\int_{\pi/2}^{\pi/3}\theta\sin 2\theta \ d\theta\\
&amp;= \int_{\pi/2}^{\pi/3}\theta\ d\cos2\theta\\
&amp;= \left[\theta\cos2\theta\right]_{\pi/2}^{\pi/3} - \int_{\pi/2}^{\pi/3}\cos2\theta\ d\theta\\
&amp;= \left[\theta\cos2\theta - \frac 12 \sin2\theta\right]_{\pi/2}^{\pi/3}\\
&amp;= -\frac{\pi}{6} - \frac{\sqrt3}{4} +\frac\pi2\\
&amp;= \frac\pi 3 - \frac{\sqrt3}{4}\\
\frac4\pi I_1 &amp;= \frac 43 - \frac{\sqrt3}{\pi}
\end{align*}$$



$$\begin{align*}
I_2 &amp;= \int_0^1 r^2\sqrt{2^2-r^2}\ dr\\
&amp;= -16\int_{\pi/2}^{\pi/3} \cos^2\theta\sin^2\theta\ d\theta&amp;&amp;(r=2\cos \theta)\\
&amp;= -4\int_{\pi/2}^{\pi/3} \sin^2 2\theta\ d\theta\\
&amp;= -4\int_{\pi/2}^{\pi/3} \frac{1-\cos4\theta}{2}\ d\theta\\
&amp;= -2\left[\theta-\frac{1}{4}\sin4\theta\right]_{\pi/2}^{\pi/3}\\
&amp;= -2\left(\frac\pi3+\frac{\sqrt3}8-\frac\pi2\right)\\
&amp;= \frac\pi3-\frac{\sqrt3}{4}\\
\frac1\pi I_2 &amp;= \frac13-\frac{\sqrt3}{4\pi}
\end{align*}$$



$$p = 1-\frac{3\sqrt3}{4\pi}$$
Unless you are given a specific sample space, the answer is 1/2, as there are uncountably many pairs of points distance r apart and uncountably many not. Without a sample space, the addition rule for probabilities no longer applies, so you have a 1/2 probability of the distance being each of less than, equal to, and greater than r.