A parabola (say $y^2=4ax$) divides the coordinate plane into two regions: one considered to be "inside" the parabola and the other "outside". However, both these regions are infinitely large. What is the probability that a randomly chosen point on this plane lies inside the parabola?
Of course, had the curve been one with finite area, one could argue that the probability is zero (i.e. it is an almost impossible event), but this seems to be much tricker.
More generally, how would one go about finding such a probability for any curve that encloses an infinitely large area?
Best Answer
An example that shows the probability need not be zero:
Choose two independent uniforms $(u_1,u_2)$ on $(0,1)$ and let $r=−\ln u_1$ and $\theta=2\pi u_2$. Then the point described in polar coordinates by $(r,\theta)$ is randomly distributed over the plane. The distribution is not uniform; the distance from the origin is exponentially distributed, thus concentrated near the origin.
Then for any given $\theta$ with $0 < \theta \leq \pi/2$, the value of $u_1$ that causes the point to lie just on the parabola is $$u_1 = e^{-\frac{\cos\theta \cot \theta}{4a}}$$ When $\pi/2 < \theta < \pi$, that quadrant (by symmetry) has the same probability of $(r,\theta)$ lying inside the parabola as in the first quadrant; when $\pi \leq \theta < 2\pi$ the point of course cannot lie inside the parabola.
So the overall probability of $(r,\theta)$ lying inside the parabola will be given by $$ \frac12 \frac2{\pi} \int_0^{\frac{\pi}{2}}e^{-\frac{\cos\theta \cot \theta}{4a}}\,d\theta $$ That integral is not zero. For example, when $a=1$ the probability of lying inside the parabola is about $0.3617$.
(Here, I did this for the upward-facing parabola $y=4ax^2$ but for your example it will work along the same general lines.)