Consider the sample space.
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline
& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
1 & – & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) & (1,7) & (1,8) & (1,9) \\ \hline
2 & (2,1) & – & (2,3) & (2,4) & (2,5) & (2,6) & (2,7) & (2,8) & (2,9) \\ \hline
3 & (3,1) & (3,2) & – & (3,4) & (3,5) & (3,6) & (3,7) & (3,8) & (3,9) \\ \hline
4 & (4,1) & (4,2) & (4,3) & – & (4,5) & (4,6) & (4,7) & (4,8) & (4,9) \\ \hline
5 & (5,1) & (5,2) & (5,3) & (5,4) & – & (5,6) & (5,7) & (5,8) & (5,9) \\ \hline
6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & – & (6,7) & (6,8) & (6,9) \\ \hline
7 & (7,1) & (7,2) & (7,3) & (7,4) & (7,5) & (7,6) & – & (7,8) & (7,9) \\ \hline
8 & (8,1) & (8,2) & (8,3) & (8,4) & (8,5) & (8,6) & (8,7) & – & (8,9) \\ \hline
9 & (9,1) & (9,2) & (9,3) & (9,4) & (9,5) & (9,6) & (9,7) & (9,8) & – \\ \hline
\end{array}$
Therefore, the required probability
$\begin{array}{cl}
= & \dfrac{16}{72} \\
= & \dfrac{2}{9}
\end{array}$
