Here is another example:
$ \mathtt{L} = \{ x\#y \mid x \in \mathtt{EQ},y \in \overline{\mathtt{EQ}} \} $,
where
$ \mathtt{EQ}=\{ a^nb^nc^n \mid n \geq 0 \} $ and $ \overline{\mathtt{EQ}} $ is the complement of $\mathtt{EQ}$.
It is a well known fact that $\mathtt{EQ}$ is not in $ \mathsf{CFL} $.
Assume that $ \mathtt{L} $ is recognized by a PDA $ \small \mathcal{P_1}$. We construct a new PDA $ \small \mathcal{P}'$. On input $w$, $ \small \mathcal{P}'$ simulates $ \small \mathcal{P}_1$ on the string $w\#a$. Since $ \small \mathcal{P}'$ clearly recognizes $ \mathtt{EQ} $, we conclude that $ \mathtt{L} \notin \mathsf{CFL}$.
Similarly, assume that the complement of $ \mathtt{L} $ is recognized by a PDA $ \small \mathcal{P}_2$. We build another PDA $ \small \mathcal{P}''$. On input $w$, $\small \mathcal{P}''$ simulates $\small \mathcal{P}_2$ on the string $\#w$. $ \small \mathcal{P}''$ also recognizes $ \mathtt{EQ} $, so $ \mathtt{L} $ can not be in $\mathsf{coCFL}$ either.
$ \mathtt{EQ}$ can be recognized by a (one-way) probabilistic one-counter automaton (P1CA) with any desired error bound (Freivalds, 1979). So, it is not hard to show that $\mathtt{L}$ can also be recognized by a P1CA with any desired error bound.