Three light bulbs

Three light bulbs with rates $\lambda_1 = 1$ , $\lambda_2 = 2$ , and $\lambda_3 = 3$ serving simultaneously form a union of three Poisson processes (the death of a light bulb can be thought of as a first event in the possibly infinite process where you replace light bulbs with new ones with the same characteristics, and we know that inter-arrival times in a Poisson process follow the exponential distribution). The probability of occurrence of any of the three independent events is the sum of their probabilities so we have a Poisson process with the rate $\lambda_P = 1 + 2 + 3$ . The first light bulb is expected to go out at time $\frac{1}{\lambda_P}$ . Then there are three possibilities: any of the $i$ light bulbs could die first with the probabilities $\frac{\lambda_i}{\lambda_1+\lambda_2+\lambda_3}$ . When this happens we are left with a new merged Poisson process with a rate equal to $\lambda_1+\lambda_2, \lambda_1+\lambda_3$ , or $\lambda_2+\lambda_3$ , and the reasoning can be repeated. The resulting formula is

$\displaystyle \frac{1}{\lambda_1+\lambda_2+\lambda_3} + \frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3} \Big( \frac{1}{\lambda_2+\lambda_3} + \frac{\lambda_2}{\lambda_2+\lambda_3}\frac{1}{\lambda_3} + \frac{\lambda_3}{\lambda_2+\lambda_3}\frac{1}{\lambda_2} \Big) + \\ \displaystyle \frac{\lambda_2}{\lambda_1+\lambda_2+\lambda_3} \Big( \frac{1}{\lambda_1+\lambda_3} + \frac{\lambda_1}{\lambda_1+\lambda_3}\frac{1}{\lambda_3} + \frac{\lambda_3}{\lambda_1+\lambda_3}\frac{1}{\lambda_1}\Big) + \frac{\lambda_3}{\lambda_1+\lambda_2+\lambda_3}\Big( \frac{1}{\lambda_1+\lambda_2} + \frac{\lambda_1}{\lambda_1+\lambda_2}\frac{1}{\lambda_2} + \frac{\lambda_2}{\lambda_1+\lambda_2}\frac{1}{\lambda_1} \Big)$

When the rates 1, 2, and 3 are plugged in it gives $\frac{73}{60}$ , and the answer is $\fbox{133}$ .