Probability

We can choose three squares in a diagonal line parallel to $BD$ in the $\triangle ABD$. Clearly, three squares in $\triangle ABD$ and in a diagonal line parallel to $BD$ can be chosen in${}^{3}C_{3}+{}^{4}C_{3}+{}^{5}C_{3}+{}^{6}C_{3}+{}^{7}C_{3}+{}^{8}C_{3}$ waysSimilarly, in $\triangle BCD$ the squares can be chosen parallel to $BD$ in an equal number of ways.Hence, the total number of ways in which three squares can be chosen in a diagonal line parallel to $BD$ is$=2({}^{3}C_{3}+{}^{4}C_{3}+{}^{5}C_{3}+{}^{6}C_{3}+{}^{7}C_{3})+{}^{8}C_{3}$( $\because BD$ is common to both the triangles) Similarly, squares can be chosen in a diagonal line parallel to $AC$ and hence the total number of favourable ways$=4({}^{3}C_{3}+{}^{4}C_{3}+{}^{5}C_{3}+{}^{6}C_{3}+{}^{7}C_{3})+2\cdot{}^{8}C_{3}=392$Hence, the required probability$=\;\frac{392}{{}^{64}C_{3}}=\;\frac{392\times6}{64\times63\times62}=\;\frac{7}{744}$