I have a queue where jobs arrive as a Poisson process with rate $\lambda=\frac{1}{6}$ jobs per minute. The process time of a job is exponentially distributed with mean time 3 minutes with probability $\frac{2}{5}$ or is the sum of two independent exponential distributions with each a mean time of 3 minutes with probability $\frac{3}{5}$. I want to compute the expected residual service time.
I know it is given by $\mathbb{E}[R]=\frac{\mathbb{E}[B^2]}{2\mathbb{E}[B]}$ where $B$ denotes the service time distribution.
So I compute $\mathbb{E}[B]=\frac{2}{5}\cdot 3+\frac{3}{5}\cdot 6=\frac{24}{5}$ and $\mathbb{E}[B^2]=\frac{2}{5}\cdot 3^{2}+\frac{3}{5}6^2=\frac{126}{5}$.
This then gives me $\mathbb{E}[R]=\frac{21}{8}$. The solution of this exercise gives that \begin{equation} \mathbb{E}[R]=\frac{5}{8}\cdot 3+\frac{3}{8}\cdot 6=\frac{33}{8} \end{equation}
I don't understand which method is used in the solution and why it would give a different result than my approach. Could someone help me by explaining what happens in the given solution and/or is wrong with my approach?