# Probability that random circle inside of unit circle contains unit circle origin

A point $$(x, y)$$ inside of a unit circle is picked uniformly at random. Then a radius $$r$$ is picked at random such that a circle $$(x, y, r)$$ is inside the unit circle. What is the probability that unit circle's origin is inside of circle $$(x, y, r)$$?

I'll assume that the unit circle is centered at the point $$(0, 0)$$.

I tried to fix the $$y$$ coordinate of the random point to $$0$$ and then try to think about all possible $$x, r$$ values. This has lead me to the following inequalities: $$0 \leq x \leq 1$$ $$0 \leq r \leq 1 - x$$

This corresponds to all radii such that a new circle is inside of unit one. Now, for a new circle to contain $$(0, 0)$$ $$x \leq r$$

must hold.

This has lead me to calculating the probability of $$\frac{1}{2}$$ but it seems to be incorrect. Could you help me please?

• So the success probability for a fixed $x$ is $(1-2x)/(1-x)$. Now integrate over $x$. – Marcus Ritt May 16 '19 at 11:43
• I think you must be thinking $x\leq r\leq 1-x$ implies $x\leq\frac12.$ True enough, but this is a necessary condition, not a sufficient one. If $x=.4$ say, $r$ has to be between $.4$ and $.6$. – saulspatz May 16 '19 at 11:51
• Don't forget to work out the distribution of $x$ within the interval $[0,1].$ It isn't uniform. – David K May 16 '19 at 12:00

First we need to find the probability of $$x,y$$ to be on distance $$R$$ from the center: CDF is $$P(d, so PDF is $$\omega(R) = 2R$$.
Assuming that “radius picked at random” means, uniform distribution from 0 to 1-R, then the probability of $$r-R>0$$: $$\int_0^{1/2} \int_R^{1-R}\omega(R)\omega(r)drdR=\int_0^{1/2} \int_R^{1-R}\frac{2R}{1-R}drdR=\int_0^{1/2}\frac{2R(1-2R)}{1-R}dR\\ = \frac32 - \log 4$$