Is $L = \{\langle M, w \rangle: \text{on input}\ w\ M\ \text{doesn't pass through a state}\ q\}$ decidable?
You can prove that $L$ is undecidable using a diagonal argument: Assume $M$ decides $L$. Then, define $\tilde{M}$ as follows: On input $x$, run $M$ on input $\langle x, x \rangle$. If it accepts, go through all states and then halt; otherwise halt without going to some dedicated state $q_0$.

If you now run $\tilde{M}$ on input $\tilde{M}$, it runs $M$ on input $\langle \tilde{M}, \tilde{M} \rangle$. Since $M$ decides $L$, it accepts if and only if $\tilde{M}$ on input $\tilde{M}$ does not go through some state. This contradicts the construction of $\tilde{M}$. Hence, $L$ is not decidable.