Ans: (a) $24$ (b) $\dfrac{2}{5}$
- Since the mean of the ten numbers is $11$, then we have
$\begin{array}{rcl}
\dfrac{2+3+5+\ldots+19+k}{10} & = & 11 \\
86 + k & = & 110 \\
k & = & 24
\end{array}$ - The required probability
$\begin{array}{cl}
= & \dfrac{4}{10} \\
= & \dfrac{2}{5}
\end{array}$
