Required probability has $2$ parts.
a) When both $1$ and $2$ remain after $10$ are chosen. Later $1$ will be chosen with probability $1$ and hence, $2$ will remain.
b) When $1$ is thrown along with some other $9$, and $2$ remains. Later we have to multiply with probability that $2$ will not be thrown.
$P$(2 remains after all the events) = $P$(1 and 2 are not thrown out) + $P$(1 is thrown out , 2 remains) * $P$(from remaining 10 2 is not chosen)
$\rightarrow$ $\frac{_{10}^{18}\textrm{C} + _{9}^{18}\textrm{C} * \frac{9}{10} }{ _{10}^{20}\textrm{C}} $
$\rightarrow \frac{9}{19}$
Hence, $B$.