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Create a NFA for the following language $$\{w \mid w \text{ contains an even number of $0$s, or contains exactly two $1$s }\} \enspace.$$ Do it with six states: My NFA

However, I think I made it too hard and the following one is far simpler.

Desired NFA

I don't understand the empty strings of the latter. If a string like 00 is passed in the latter, it wouldn't read it to me because it would get stuck in the first state. There is no edge labeled 0 going out in the first state.

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    NFA always sounds like some expletive ridden acronym to me. Ignore the NFA downvoters. – 2017-02-05

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The $\epsilon$ that labels the transitions out of the initial state "consumes no input." That is, the NFA can move to either successor of the initial state before reading the first letter of the input word.

A way to interpret what happens is this NFA is that the automaton chooses nondeterministically whether to look for an even number of zeros (upper branch) or exactly two ones (lower branch) before it reads the input.

(Formally, this NFA has one or two runs for each word. If at least one run is accepting, the automaton accepts the word.)

It is not essential that the initial state be accepting, but it makes the elimination of the $\epsilon$-transitions marginally easier, because the initial state after removing $\epsilon$-transitions is accepting.

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    that makes this very easy to do cause I just have to do it one by one and merge them with the empty strings. – 2017-02-05
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    Yes, that's the beauty of $\epsilon$-transitions. They take care of "either this or that" quite naturally. – 2017-02-05