Consider the automata given in previous question. The minimum state automaton…

The automaton has three states: q0, q1, and q2. State q3 is unreachable from the start state q0 because there is no path to it, so it can be removed. After removing q3, the remaining states are q0, q1, and q2. State q1 is a final state (double circle), while q2 is not, so they are distinguishable. The transitions from q0 on 'a' go to q1 (final), and from q1 and q2, they form a cycle on 'a' and 'b'. By minimizing the DFA using standard state equivalence techniques, we find that q1 and q2 are not equivalent due to their final status. However, the minimal DFA can be reduced to 2 states by merging indistinguishable states. The correct answer is 2.