Triangle in a circle (vs triangle)

Suppose $P$ and $Q$ placed randomly. Join the points $P$ and $O$ and extend it to the point $A$ on the circumference. Do the similar to the segment $QOB$
If the triangle $PQR$ contains the point $O$ , then the final point $R$ must be placed within the minor arc $AB$ colored red (as shown). So the probability, $x$ is the length of minor arc $AB$ divided by the circumference. Suppose the circumference is of length 1 unit. Then $x$ is the length of minor arc $AB$ , which is the same as the length of minor arc $PQ$ .

Now fix the point $P$ and draw the segment $POA$ . By symmetry, the point $Q$ is either above $PA$ or below $PA$ . Without loss of generality, let the point $Q$ be above $PA$ . As $Q$ is chosen uniformly at random on the arc $PA$ , averagely the point $Q$ will fall on the center of the arc $PA$ , which means that the length of minor arc $PQ$ is $\frac{1}{4}$ , and so the length of minor arc $AB$ . This implies that $x=\frac{1}{4}$ and hence $\displaystyle\left\lfloor 1000x \right\rfloor =250$ .

A very similar problem was posts about one month ago . I'll copy the relevant portion of my solution below.

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Given two points $A, B$ on a circle, the third point $C$ must lie in the red region below if $\triangle ABC$ is to contain the circle's center:

Now, we find the probability of this happening. Fix $A$ and $B$ on the circle, and let $\theta$ be the angle between them ( $\angle AOB$ ). Then $0<\theta\leq \pi$ . Note that the angle intercepting the red region in the diagram is precisely $\theta$ . Thus the probability of a random point $C$ being placed in the red area is $\frac{\theta}{2\pi}$ . We integrate to compute the average probability over all possible $\theta$ : $\frac{1}{\pi-0}\int_0^\pi \frac{\theta}{2\pi}\,\mathrm{d}\theta = \frac{1}{\pi}\left[\frac{\theta^2}{4\pi}\right]_{\theta=0}^\pi = \frac{\pi^2}{4\pi^2} = \frac{1}{4}$

$\left\lfloor 1000\times\frac{1}{4} \right\rfloor = \left\lfloor 250 \right\rfloor = \boxed{250}$

Probability is simply the probability of an acute triangle from 3 random points on a circle.

$=\frac {1}{π-0}\int ^{π}_{0}\frac {\theta }{2π}d\theta = 0.25$