Let $T_1, T_2$ be exponential random variables with parameter $\lambda$ and $S$ an exponential random variable with parameter $\mu$. It is assumed also that these are all independent. We are asked to calculate the expected value $\mathbb{E}[\min(T_1+T_2, S)]$.
I got the wrong answer by solving it like this:
Consider the given random variables as arrival times of the first arrival in Poisson processes. Then merge the processes of $T_1$ and $T_2$. That process then has parameter $2\lambda$, right? Call that process $M$.
Then merge $M$ with the process of $S$.
Now we must solve the expected value of the random variable
$$X = \text{time before arrival from $S$ or two arrivals from $M$}.$$ I started solving it by conditioning on whether the first arrival is from $M$ or $S$: $$\mathbb{E}[X] = \mathbb{P}(1\text{st arrival is from M})\mathbb{E}[X\space | \space 1\text{st arrival is from M}] + \mathbb{P}(1\text{st arrival is from S})\mathbb{E}[X\space | \space 1\text{st arrival is from S}]$$
$$=\frac{2\lambda}{\mu + 2\lambda}\mathbb{E}[X\space | \space 1\text{st arrival is from M}] + \frac{\mu}{\mu + 2\lambda}\frac{1}{\mu}.$$ Here I'm not sure if this is correct but for calculating $\mathbb{E}[X\space | \space 1\text{st arrival is from M}] $ we have the first arrival given that it comes from $M$ so add the expected for that and start the process anew after the that first arrival and condition similarly on where the second arrival comes from to get
$$\mathbb{E}[X\space | \space 1\text{st arrival is from M}] $$
$$= \frac{1}{2\lambda} + \frac{\mu}{\mu+2\lambda}\frac{1}{\mu} + \frac{2\lambda}{\mu+2\lambda}\frac{1}{2\lambda}.$$
So alltogether I got
$$\mathbb{E}[X] = \frac{2\mu+8\lambda}{(\mu+2\lambda)^2}.$$
But like I said this is not correct (simulation doesn't support this and the answer should actually be $\frac{\mu+2\lambda}{(\mu+2\lambda)^2}$. The question and a solution are from here, problem set 7, question 3. I don't understand the solution given there (why is there no $2\lambda$?) and I'd like to know what's wrong with my answer.