Does a NFA/DFA need a failure state?
A state in a DFA always needs to have exactly one transition per symbol in the alphabet. I have understood from a few of my classmates that some teachers omit the error-state. I however prefer to show the error-state in graphs, since it simplifies verification of determinism in a graph.

In a NFA there can be no such rule. Therefore it should work to not present a transition for the symbol (0) in this case. If the NFA would be determinized, this non existing transition, along with all other non-looping transitions would point to a error state.