Just like Gianlino's Disc. Only 1000 times easier

First thing to note, is that the circumcenter is situated on the y-axis, regardless of where point $E$ is at. This is so because the circumcenter is a point which the perpendicular bisectors of $AB$ , $AE$ and $EB$ will intersect, and the perpendicular bisector of $AB$ would just be the line $x=0$ .

Due to symmetry, we just have to focus on the first quadrant of the circle. Let $O$ be the origin, and $\angle EOA=a$ . Now, when $a$ is smaller than a certain value $k°$ , the circumcenter will be out of the circle of radius 3. When $a$ is bigger however, but not more than 90°, the circumcenter will be within the circle. So what we need to do is find that value of $k$ . To do that, we make the circumcenter situate at $(0,3)$

From the fact that the (black) circle's center is $(0,3)$ , and it passes through $(2,0)$ , we can find, through Pythagoras theorem, the radius of the circle as $\sqrt{13}$ . Redrawing the diagram, we get

Through this, we can find the value of $k$ using the cosine rule, which happens to be $30$ .

Therefore, if $0°<\angle EOA < 30°$ , the circumcenter would be out of the circle of radius $3$ . If $90°>\angle EOA > 30°$ , the circumcenter would be in the circle.

Hence, the probability that the circumcenter is in the circle or radius $3$ is $\frac{90-30}{90}=\frac{2}{3}$

But wait... then the answer should be $5$ ...

This simulator might help.

Consider E As (cosx , sinx)

It is easy to observe that circumcentre lies on the line x=0

Now just write the equation of perpendicular bisector of one side in terms cosx and sinx and calculate y coordinate.

That will be -3/2 sinx

Just its modulus should be less than 3 to lie in interior of circle