Ans: (a) $\dfrac{5}{18}$ (b) $\dfrac{67}{90}$
- The required probability
$\begin{array}{cl}
= & \dfrac{C^5_2}{C^9_2} \\
= & \dfrac{5}{18}
\end{array}$ - The required probability
$\begin{array}{cl}
= & \dfrac{5}{18} \times \dfrac{C^{10}_3}{C^{10}_3} +\dfrac{C^5_1C^4_1}{C^9_2} \times \dfrac{C^9_3}{C^{10}_3} +\dfrac{C^4_2}{C^9_2} \times \dfrac{C^8_3}{C^{10}_3} \\
= & \dfrac{67}{60}
\end{array}$
