Ans: A
Note that the sample space is
Note that the sample space is
$\begin{array}{|c|c|c|c|c|c|c|c|} \hline
& 1 & 1 & 1 & 2 & 2 & 3 & 4 \\ \hline
1 & – & 2 & 2 & 3 & 3 & 3 & 5 \\ \hline
1 & 2 & – & 2 & 3 & 3 & 3 & 5 \\ \hline
1 & 2 & 2 & – & 3 & 3 & 3 & 5 \\ \hline
2 & 3 & 3 & 3 & – & 4 & 5 & 6 \\ \hline
2 & 3 & 3 & 3 & 4 & – & 5 & 6 \\ \hline
3 & 4 & 4 & 4 & 5 & 5 & – & 7 \\ \hline
4 & 5 & 5 & 5 & 6 & 6 & 7 & – \\ \hline
\end{array}$
& 1 & 1 & 1 & 2 & 2 & 3 & 4 \\ \hline
1 & – & 2 & 2 & 3 & 3 & 3 & 5 \\ \hline
1 & 2 & – & 2 & 3 & 3 & 3 & 5 \\ \hline
1 & 2 & 2 & – & 3 & 3 & 3 & 5 \\ \hline
2 & 3 & 3 & 3 & – & 4 & 5 & 6 \\ \hline
2 & 3 & 3 & 3 & 4 & – & 5 & 6 \\ \hline
3 & 4 & 4 & 4 & 5 & 5 & – & 7 \\ \hline
4 & 5 & 5 & 5 & 6 & 6 & 7 & – \\ \hline
\end{array}$
Therefore, the required probability
$\begin{array}{cl}
= & \dfrac{10}{42} \\
= & \dfrac{5}{21}
\end{array}$
