M/M/1 queue with two types of customers, distribution of the total number of customers

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I have problems to derive the following:

"A gas station offers two services. For each service customers arrive according to a Poisson process. On average 20 customers per hour for service 1 and 5 customers for service 2. The service times are exponential, both with a mean service time of two minutes.

Determine the distribution of the total number of customers in the system."

Using minutes as time units, I get $\rho_1=\lambda_1 \cdot \mathbb{E}[B]= \frac{1}{3} \cdot 2 = \frac{2}{3}$ and $\rho_2=\lambda_2 \cdot \mathbb{E}[B]= \frac{1}{12} \cdot 2 = \frac{1}{6}$. So the distribution of the number of customers looking at the different services individually is $$\mathbb{P}(L_1=n)=(1-\rho_1)(\rho_1)^n=\frac{1}{3}\Big(\frac{2}{3}\Big)^n$$ and $$\mathbb{P}(L_2=n)=(1-\rho_2)(\rho_2)^n=\frac{5}{6}\Big(\frac{1}{6}\Big)^n .$$

For the distibution of the total number of customers, I thought about $$\mathbb{P}(L=n)=\sum_{k=0}^{n}\mathbb{P}(L_1=k,L_2=n-k).$$

From there, I do not really know how to compute is further. Am I on the right track? Does anybody has a hint how to compute it?