Answer

Start with the merged Poisson process which will denote the time till the first bulb will fail. For this process, \(\lambda^{\prime} = 3\lambda\). Hence, \(E[\text{first bulb fails}] = \frac{1}{3\lambda}\). After the first bulb dies out, we are left with a process with \(\lambda^{\prime} = 3\lambda\). Due to memoryless property, \(E[\text{second bulb fails}] = \frac{1}{2\lambda}\) and consequently \(E[\text{last bulb fails}] = \frac{1}{\lambda}\).

Note the above two times denote the time difference, i.e. the time taken for the bulb to die out after the last bulb died out. Thus, \(E[\text{time until last bulb dies out}] = \frac{1}{3\lambda} + \frac{1}{2\lambda} + \frac{1}{\lambda}\)