What is the probability of a random line passing through the unit disk meet the y-axis? This is a question that I came up myself. So it might have some of the errors pointed in the comments below.
Consider the open unit disk centered at (0,0) and the set of all straight lines of the plane that intersect the y-axis. Hence, we are consedering the set of straight lines that have the form: y=lx+m, $l\in\mathbb{R}, m\in\mathbb{R}$. What is the probability of a randomly selected line that passes through the unit disk to intesect the y-axis inside the unit disk?
I don't really know how to approach this kind of problems. At first I try to see which lines pass through the unit disk. The coefficients must obey $|m|<\sqrt{l^2+1}$. To intersect the y-axis inside the unit disk we must have $|m|<1$. Thus, for a fixed $l\in\mathbb{R}$ the proportion of the lines that meet the y-axis inside the unit disk is $1/\sqrt{l^2+1}$ (lenght of (-1,1)/ length of $(-\sqrt(l^2+1),\sqrt(l^2+1))$. Now the probability must be the mean value of $1/\sqrt{l^2+1}$
for $l\in(-\infty,+\infty)$, which is the integral of this function over that interval. But the integral
$$\int\limits_{-\infty}^{\infty}\dfrac{1}{\sqrt{l^2+1}}=\infty.$$
Hence, this reasoning - which seems right to me - doesn't lead to an answer. I don't really know how to approach this kind of problems.
Do you have any hints? Suggestions? Or maybe an answer?
 A: It's better to think of the lines in terms of their angle of approach. Fundamentally the probability of a line intersecting the $y$ axis is the area of the central four sided figure swept out by sliding a particularly angled line left or right until it's intersection with the axis lies outside the disk.
The area of a segment of a circle i.e. the "cap" of a sector cut off by a chord is given by
$$\frac{\beta-\sin\beta}{2}$$
where $\beta$ is the angle substended by the arc as measured from the center of the circle. This angle and the angle of attack of the lines (as measured from the $y$ axis) form an equilateral triangle giving the area of that segment as
$$\frac{\pi-2\theta - \sin2\theta}{2}$$
Since there are two of these segments outside the area swept out by the circle, the probability we want is the area of the full circle minus twice this segment.
$$P(\theta) = \frac{2\theta+\sin2\theta}{\pi}$$
where the resultant is normalized by the total area. This formula is only valid until the angle reaches $\pi/2$, at which point the distribution is symmetric and will retreat back to $0$ towards $\pi$.
Using $\Theta \sim U(0,\pi)$ we can now compute the integral
$$\frac{1}{\pi}\int_{[0,\pi]}P(\theta)d\theta = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \frac{2\theta+\sin2\theta}{\pi}d\theta = \frac{1}{2}+\frac{1}{\pi^2}$$
