Going from an M/M/1/∞ to an M/M/1/N queue If I initially have an M/M/1/∞ queue with arrival rate of 6/hour (people arrive according to a Poisson distribution) and service rate 5/hour (average service time is exponentially distributed). But I want to reduce the arrival rate to about 4/hour and to do so I provide a limited number of seats in a waiting room, so now I have an M/M/1/N queue with arrival rate of about 4 and service rate 5/hour. In the new queueing system, when all seats are taken, new arriving customers go elsewhere.
I want to determine what N should be in order to achieve an arrival rate of about 4.
What I thought of doing is, taking different values of λ (e.g. 5,4,3) and calculating the proportion of lost customers when switching from the M/M/1/∞ to and M/M/1/N, which equals to λ times the probability that there are N customers in the system, for each value of λ.
Then I would get different values of N for different λ. 
Is this the right way to solve this question?
Thanks in advance!

Since you have an arrival rate of $6/$hour and you need an arrival rate of $4/$hour you need to compute a probability $p$ of accepting a customer to the system, such that $6p=4\implies p=2/3$. The probability $1-p=\frac13$, which is the probabitily of rejecting a customer, is equal to the probability that an arriving customer sees $N$ customers in the system (waiting seats and service), i.e. equal to $p_{N}$. Hence, you need $N$ such that $$p_{N}=1-\frac23=\frac13$$ Letting $\rho=λ/μ=6/5$ you have that $$p_{N}=\frac{ρ^{N}(1 − ρ)}{1 − ρ^{N+1}}
\iff \frac13=\frac{(6/5)^{N}(-1/5)}{1-(6/5)^{N+1}}\iff N=\frac{\ln{(5/3)}}{\ln(6/5)}\approx 2.8$$ and hence $N$ must me an integer, you get $N=3$.
