Correct. The number of ways 10 robots can finish in the top three spots is $\displaystyle\frac{{{10}}}{{{73}}}={120}\usepackage{cmbright}\Huge\$ $\displaystyle{\left({n}={10}{\quad\text{and}\quad}{k}={3}\right)}\usepackage{cmbright}\Huge\$ . The number of ways that two of the three robots can finish in the top three spots along with a different robot is $\displaystyle\frac{{{7}}}{{{61}}}={7}\usepackage{cmbright}\Huge\$ . The probability is the ratio of $\displaystyle\frac{{7}}{{120}}\usepackage{cmbright}\Huge\$ .