Ans: (a) $\dfrac{54}{133}$ (b) $\dfrac{79}{133}$
- The required probability
$\begin{array}{cl}
= & \dfrac{C^{10}_2 C^{12}_2}{C^{22}_4} \\
= & \dfrac{54}{133}
\end{array}$ - The required probability
$\begin{array}{cl}
= & 1 -\dfrac{54}{133} \\
= & \dfrac{79}{133}
\end{array}$
