
$\begin{aligned}
& \text { Total }={ }^{16} \mathrm{C}_2 \
& \text { Required ways }=\text { Total }- \text { (adjacent square) } \
& ={ }^{16} \mathrm{C}_2- \
& \text { [3 pair in vertical & horizontal for } \
& \text { each row and column] } \
& ={ }^{16} C_2-[3 \times 4+3 \times 4] \
& =96 \
& \text { Probability }=\frac{96}{120}=\frac{4}{5}
\end{aligned}$
Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :