Consider the following deterministic finite automaton (DFA) defined over the…

Final answer: The DFA accepts exactly the set of all strings that contain an even number of b's.

Reasoning: the automaton effectively tracks the parity of the number of b's seen so far.

• Start/accepting state = even number of b's observed so far (including zero).

• Reading an a does not change parity: transitions on a stay within the same parity class, so a's do not affect acceptance.

• Reading a b toggles parity: each b moves the machine between the even and odd parity states. Therefore acceptance depends solely on whether the total number of b's is even.

Examples (to build intuition):

• Accepted: the empty string (0 b's), "aa" (0 b's), "bb" (2 b's), "abaabb" (2 b's).

• Rejected: "b" (1 b), "ab" (1 b), "bba" (1 b), "bbab" (3 b's).

Why the other described languages do not match the DFA:

• The property "contains the substring 'bab'": not equivalent. Counterexample: "bbab" contains 'bab' but has three b's and is rejected.

• The property "ends with 'bab'": not equivalent. Counterexample: "bbab" ends with 'bab' but is rejected since it has an odd number of b's.

• The property "does not contain 'aba'": not equivalent. Counterexample: "b" does not contain 'aba' but is rejected because it has an odd number of b's.

Conclusion: The DFA accepts exactly those strings over {a, b} whose total number of b's is even.