Final answer:
If bulb c burns out in the given circuit, bulb b will turn off while bulb a will remain lit.
Step-by-step explanation:
In the given circuit diagram, if bulb c burns out, bulb b will turn off. When bulb c burns out, it breaks the circuit and interrupts the flow of current. As a result, no current can reach bulb b, causing it to turn off. However, bulb a will still remain lit because it is connected separately and is not affected by the burnout of bulb c.