Ans: B
Note that the sample space is
Note that the sample space is
$\begin{array}{|c|c|c|c|c|} \hline
& 3 & 5 & 7 & 9 \\ \hline
3 & – & 15 & 21 & 27 \\ \hline
5 & 15 & -& 35 & 45 \\ \hline
7 & 21 & 35 & – & 63 \\ \hline
9 & 27 & 45 & 63 & – \\ \hline
\end{array}$
Therefore, the required probability
$\begin{array}{cl}
= & \dfrac{4}{12} \\
= & \dfrac{1}{3}
\end{array}$
