To solve this problem, we need to analyze the M/M/1 queue system and verify each of the given options. An M/M/1 queue is characterized by a single server with Poisson arrivals and exponentially distributed service times, which are independent of each other.

The arrival rate (λ) is given as \(\lambda = 3\), and the service rate (μ) can be derived from the mean service time, which is \(\frac{1}{9}\). Therefore, the service rate (μ) is:

\(\mu = \frac{1}{(\text{mean service time})} = 9\)

In an M/M/1 queue, the traffic intensity (ρ) is:

\(\rho = \frac{\lambda}{\mu} = \frac{3}{9} = \frac{1}{3}\)

The long run probability that there are exactly \(n\) customers in the system, \(P_n\), is given by:

\(P_n = (1-\rho) \rho^n\)

Let's evaluate each option:

• \(P_0 = (1-\rho) = \frac{2}{3}\). This is not equal to \(\frac{1}{3}\), so this option is incorrect.
• \(P_1 = (1-\rho) \rho = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\). This matches the given option, so it is correct.

The average number of customers in the system, \(L\), is given by:

\(L = \frac{\rho}{1-\rho} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}\)

Thus, the statement that the average number of customers is 1 is incorrect.

The average amount of time that a customer spends in the system, \(W\), is:

\(W = \frac{1}{\mu - \lambda} = \frac{1}{9 - 3} = \frac{1}{6}\)

This matches the given option, so it is correct.

Therefore, the correct options are \(P_1 = \frac{2}{9}\) and the average amount of time that a customer spends in the system is \(\frac{1}{6}\).