How to calculate a capacitance of a probe in a equivalent circuit of oscilloscope?
\$\frac{1}{10} \$ ratio is set by the voltage divider:

\$ H(s=0)=\frac{R_{osc}}{R_{osc}+R_{probe}}\$

In order to get a flat attenuation the following condition has to be verified:
\$ R_{probe}C_{probe}=R_{osc}(C_{osc}+C_{cable}) \$ that means:
\$ C_{probe}=\frac{R_{osc}(C_{osc}+C_{cable})}{R_{probe}}\$