Suppose a circle with the radius R. We randomly select a point within or on the circle. What is the probability of the chosen point distance from the center of the circle be smaller or equal to R/2?
I have two approaches as below:
Approach 1: The chosen point distance from the center is smaller or equal to R/2, if the point is within or on the concentric circle with the radius of R/2. The probability of randomly chosen point a point been in that area is $\pi*(R/2)^2\;\div\;\pi*R^2=1/4$
Approach 2: We consider a random radius and then pick a point on this radius. Now the chance of picking a point located on the inner half of the radius (between the center and R/2) is 1/2.
Could someone explain this?
Best Answer
Let $R$ be the radius of the original circle. If randomly selecting a point from the circle implies that every point in the circle has the same (relative) probability of being picked, then the first approach is correct. That is, the probability of a random point being within the circle of radius $R/2$ is $$\frac{\pi (R/2)^2}{\pi R^2} = \frac{1}{4}.$$
The reason the second approach doesn't work is that there is a higher likelihood of picking points outside of the smaller circle with radius $R/2$ because of the difference in area. The concept of uniformly picking a random point $p \in (0,R)$ and then extending the result to an area doesn't work because area depends on the radius squared. This is why if you double the radius of the circle, the area does not double, but rather quadruples.
Now, we can 'fix' your second approach as follows:
We need a density function in the interval $[0,R]$ that is not uniform, but is rather an increasing function of the radius. We can start by considering the circumference of a circle or radius $r$: $$f(r) = 2 \pi r.$$ Now notice $$\int_{- \infty}^\infty f(r) dr = \int_{0}^R 2 \pi r dr = \pi R^2$$ so that the area $A= \pi R^2$of the bigger circle is the normalizing constant for the density. That is, $$f(r) = \frac{2 \pi r}{\pi R^2} = \frac{2r}{R^2} \tag{1}$$ corresponds to a valid density function defined for $r \in [0,R].$ Using $(1)$ as the new density, we obtain that the probability of a random point falling within the circle of radius $R/2$ corresponds to $$\int_0^{R/2} f(r) dr = \frac{2}{R^2} \int_0^{R/2} r dr = \frac{2}{R^2}\cdot \frac{R^2/4}{2} = \frac{1}{4}.$$ This last result is essentially the same as the first approach.