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The following question has been answered on MSE before and is indeed quite famous. I have one way of solving it and it seems quite wrong. Could anyone please help me correct it?

Consider points $A$ and $B$ on the circle such that they subtend an angle $\theta$. For the centre to lie inside the triangle, $C$ should belong to the arc that is antipodal to arc ${AB}$. The probability will be $\frac{\theta}{2\pi}$. Integrating $\theta$ doesn't yield the probability. It seems like the fixing of the point $A$ is causing the issue but I don't know what to do otherwise. I would appreciate any help regarding this.

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WLOG you may fix point $A$.

If $\Theta$ denotes the length of (shortest) arc $AB$ then $\Theta$ has uniform distribution on $[0,\pi]$.

If $E$ denotes the event that triangle $ABC$ will contain the center of the circle then:$$P(E)=\frac1{\pi}\int_0^\pi P(E\mid\Theta=\theta)d\theta=\frac1{\pi}\int_0^\pi\frac{\theta}{2\pi}d\theta=\frac1{\pi}\left[\frac{\theta^2}{4\pi}\right]^\pi_0=\frac14$$ which is correct.

Compare with your own result.

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  • $\begingroup$ Why the $\frac{1}{\pi}$ at the beginning? $\endgroup$ – Mathejunior Jun 15 '19 at 7:42
  • $\begingroup$ Because the PDF of uniform distribution on $[0,\pi]$ is $\frac1{\pi}$ on interval $[0,\pi]$. $\endgroup$ – drhab Jun 15 '19 at 7:46

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