A cafeteria serving line has a coffee urn from which customers serve themselves. Arrivals at the urn follow a Poisson distribution at the rate of three per minute. In serving themselves, customers take about 15 seconds, exponentially distributed. What is the probability that three or more people are in the cafeteria?


To solve this problem, we will treat it as a **queuing problem** based on the **M/M/1 queue** model. This model assumes:

- **M** (Markovian/Poisson arrivals): The arrival of customers follows a Poisson process with a rate \(\lambda = 3\) customers per minute.
- **M** (Markovian/exponential service times): The time to serve customers follows an exponential distribution with an average service rate of \(\mu = 4\) customers per minute (since customers take about 15 seconds on average, or \(\frac{1}{4}\) minutes per customer).
- **1** (Single server): There is one coffee urn, implying only one customer can be served at a time.

### Step 1: Define the parameters
- **Arrival rate (\(\lambda\))**: 3 customers per minute.
- **Service rate (\(\mu\))**: 4 customers per minute.

### Step 2: Calculate the traffic intensity (\(\rho\))
The traffic intensity or utilization factor \(\rho\) is the ratio of the arrival rate to the service rate:

\[
\rho = \frac{\lambda}{\mu} = \frac{3}{4} = 0.75
\]

This means the server (coffee urn) is busy 75% of the time.

### Step 3: Find the probability of 3 or more customers in the system
The probability of having \(n\) customers in the system (both being served and waiting) in an M/M/1 queue is given by the formula:

\[
P(n) = (1 - \rho)\rho^n
\]

To find the probability of having 3 or more people in the system, we need to find the cumulative probability of 0, 1, and 2 people in the system, and subtract this from 1.

\[
P(\text{3 or more}) = 1 - P(0) - P(1) - P(2)
\]

Where:
- \(P(0) = 1 - \rho = 1 - 0.75 = 0.25\)
- \(P(1) = (1 - \rho)\rho^1 = 0.25 \times 0.75 = 0.1875\)
- \(P(2) = (1 - \rho)\rho^2 = 0.25 \times (0.75)^2 = 0.25 \times 0.5625 = 0.140625\)

Now, calculate the cumulative probability of 0, 1, and 2 people in the system:

\[
P(0) + P(1) + P(2) = 0.25 + 0.1875 + 0.140625 = 0.578125
\]

Finally, the probability of having 3 or more people in the system is:

\[
P(\text{3 or more}) = 1 - 0.578125 = 0.421875
\]

### Final Answer:
The probability that there are 3 or more people in the cafeteria system is approximately **0.4219** or **42.19%**.

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Would you like further explanation on this, or do you have any questions?

Here are 5 related questions to explore:
1. What is the probability that there are exactly 2 people in the system?
2. How would the result change if the arrival rate increased to 5 customers per minute?
3. What is the expected number of people in the system at any time?
4. What is the average time a customer spends in the system?
5. How does the utilization factor affect the system performance?

**Tip**: In an M/M/1 queue, the closer the traffic intensity \(\rho\) is to 1, the more congested the system becomes, increasing waiting times.

A cafeteria serving line has a coffee urn from which customers serve themselves. Arrivals at the urn follow a Poisson distribution at the rate of three per minute. In serving themselves, customers take about 15 seconds, exponentially distributed. What is the probability that three or more people are in the cafeteria?

Learn how to calculate the probability of three or more people in a cafeteria serving line using the M/M/1 queue model. This problem involves Poisson arrival rates and exponential service times. Suitable for undergraduate students studying queuing theory and probability distributions.

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