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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?

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  • $\begingroup$ So the success probability for a fixed $x$ is $(1-2x)/(1-x)$. Now integrate over $x$. $\endgroup$ – Marcus Ritt May 16 '19 at 11:43
  • $\begingroup$ 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 $.6$. $\endgroup$ – saulspatz May 16 '19 at 11:51
  • $\begingroup$ Don't forget to work out the distribution of $x$ within the interval $[0,1].$ It isn't uniform. $\endgroup$ – David K May 16 '19 at 12:00
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First we need to find the probability of $x,y$ to be on distance $R$ from the center: CDF is $P(d<R) = R^2$, so PDF is $\omega(R) = 2R$.

Assuming that “radius picked at random” means, uniform distribution from 0 to 1-R, then the probability of $r-R>0$: $$ \int_0^{1/2} \int_R^{1-R}\omega(R)\omega(r)drdR=\int_0^{1/2} \int_R^{1-R}\frac{2R}{1-R}drdR=\int_0^{1/2}\frac{2R(1-2R)}{1-R}dR\\ = \frac32 - \log 4 $$

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