Correct. The number of ways 8 robots can finish in the top three spots is $\displaystyle\frac{{{8}!}}{{{5}!{3}!}}={56}\usepackage{cmbright}\Huge\$ (n = 8 and k = 3). 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{{{5}!}}{{{4}!{1}!}}={5}\usepackage{cmbright}\Huge\$ . The probability is the ratio of $\displaystyle\frac{{5}}{{56}}\usepackage{cmbright}\Huge\$ .