Let p, q be chosen one by one from the set $\{1,\sqrt{2},\sqrt{3},2,\mathrm{e},\mathrm{\pi }\}$ with replacement. Now a circle is drawn taking (p, q) as its centre. Then the probability that at the most two rational points exist on the circle is (rational points are those points whose both the coordinates are rational)

Suppose, there exist three rational points or more on the circle

${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$. Therefore, if $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$, and $\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ are those three points, then

$\begin{array}{r}{\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{gx}}_1+2{\mathrm{fy}}_1+\mathrm{c}=0 \left(1\right)\\ {\mathrm{x}}_2^2+{\mathrm{y}}_2^2+2{\mathrm{gx}}_2+2{\mathrm{fy}}_1+\mathrm{c}=0 \left(2\right)\\ {\mathrm{x}}_3^2+{\mathrm{y}}_3^2+2{\mathrm{gx}}_3+2{\mathrm{fy}}_3+\mathrm{c}=0 \left(3\right)\end{array}$

Solving Eqs.(1), (2) and (3), we will get g, f, c as rational. Thus, center of the circle (-g, -f) is a rational point. Therefore, both the coordinates of the center are rational numbers. Obviously, the possible values of p are 1, 2. Similarly, the possible values of q are 1, 2. Thus, for this case, (p, q) may be chosen in 2 x 2, i.e., 4 ways. Now, (p, q) can be, without restriction, chosen in 6 x 6, i.e., 36 ways.

Hence, the probability that at the most two rational points exist on the circle is $\frac{36-4}{36}=\frac{32}{36}=\frac{8}{9}$