Detailed reading passage outlining the life, major periods, and artistic contributions of Pablo Picasso.

It looks like you've uploaded a text-based problem involving Poisson processes and queueing theory, specifically related to a call center scenario. Let's go through the problem carefully, solve it step by step, and explain the solution.

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🔍 Problem Summary:

We are given a Poisson process model for customer arrivals at a call center:

- Customers arrive according to a Poisson process with rate λ = 10 per hour.
- Each call has an exponentially distributed service time with mean 10 minutes → so service rate μ = 6 per hour (since 60 min / 10 min = 6 calls/hour).
- There are 3 operators (servers) in the system.
- If all operators are busy when a call arrives, the call is rejected (no waiting queue).

This is a M/M/3/3 queueing system:
- M: Markovian (Poisson arrivals)
- M: Markovian (exponential service times)
- 3: Number of servers
- 3: System capacity (no queue; maximum 3 customers in the system)

We are asked to compute:

1. The probability that a customer will be rejected (i.e., all operators busy).
2. The average number of busy operators.
3. The expected time a customer spends in the system, given that they are not rejected.

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## ✔ Step-by-step Solution

Let’s denote:
- Arrival rate: $ \lambda = 10 $ per hour
- Service rate per operator: $ \mu = 6 $ per hour
- Number of servers: $ s = 3 $
- System capacity: $ N = 3 $

This is a finite-capacity loss system (Erlang B model), where blocked calls are lost.

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📌 1. Probability that a customer is rejected

This is the blocking probability, which in an M/M/s/s system is given by the Erlang B formula:

$$
P_{\text{reject}} = P_3 = \frac{\frac{(\lambda / \mu)^3}{3!}}{\sum_{k=0}^{3} \frac{(\lambda / \mu)^k}{k!}}
$$

Let’s compute:

First, define $ a = \lambda / \mu = 10 / 6 = \frac{5}{3} \approx 1.6667 $

Now compute each term:

- $ \frac{a^0}{0!} = 1 $
- $ \frac{a^1}{1!} = 1.6667 $
- $ \frac{a^2}{2!} = \frac{(1.6667)^2}{2} = \frac{2.7778}{2} = 1.3889 $
- $ \frac{a^3}{3!} = \frac{(1.6667)^3}{6} = \frac{4.6296}{6} \approx 0.7716 $

Sum of denominator:
$$
1 + 1.6667 + 1.3889 + 0.7716 = 4.8272
$$

So,
$$
P_{\text{reject}} = \frac{0.7716}{4.8272} \approx 0.160
$$

✔ Answer 1: The probability that a customer is rejected is approximately 0.160 or 16.0%.

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📌 2. Average number of busy operators

In an M/M/s/s system, the expected number of busy servers is:

$$
L_s = \sum_{k=0}^{s} k \cdot P_k
$$

Where $ P_k $ is the probability that $ k $ servers are busy.

We already have:

- $ P_0 = \frac{1}{4.8272} \approx 0.2071 $
- $ P_1 = \frac{1.6667}{4.8272} \approx 0.3452 $
- $ P_2 = \frac{1.3889}{4.8272} \approx 0.2878 $
- $ P_3 = \frac{0.7716}{4.8272} \approx 0.1600 $

Now compute:

$$
L_s = 0 \cdot P_0 + 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3
$$
$$
= 0 + 1 \cdot 0.3452 + 2 \cdot 0.2878 + 3 \cdot 0.1600
$$
$$
= 0.3452 + 0.5756 + 0.4800 = 1.4008
$$

✔ Answer 2: The average number of busy operators is approximately 1.40.

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📌 3. Expected time a customer spends in the system, given they are not rejected

This is the conditional expected sojourn time for accepted customers.

Since there is no queue, the only time spent in the system is service time, but only if the call is accepted.

But note: All accepted customers get served immediately, since they find at least one free operator.

So, the service time is exponential with mean $ 1/\mu = 1/6 $ hours ≈ 10 minutes, regardless of how many servers are busy — because once admitted, the customer gets full service.

However, the key point is: we are conditioning on the call being accepted.

But since service time is memoryless, and accepted customers are served immediately, the time spent in the system is just the service time, which is exponentially distributed with mean $ 1/\mu = 1/6 $ hours.

Therefore, the expected time in the system for accepted customers is:

$$
W = \frac{1}{\mu} = \frac{1}{6} \text{ hours} = 10 \text{ minutes}
$$

Wait — is this correct?

Yes, because:
- No queue → no waiting time.
- Once accepted, service starts immediately.
- Service time is exponential with mean 10 minutes.

So, regardless of the state, the time in system for accepted calls is just the service time.

✔ Answer 3: The expected time a customer spends in the system, given they are not rejected, is 10 minutes.

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## ✔ Final Answers:

1. Probability of rejection: $ \boxed{0.160} $
2. Average number of busy operators: $ \boxed{1.40} $
3. Expected time in system (for accepted calls): $ \boxed{10 \text{ minutes}} $

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💡 Additional Notes:

- This is a classic Erlang B model used in telephony and call centers.
- The blocking probability increases with traffic intensity $ a = \lambda / \mu $.
- With 3 servers and $ a = 5/3 \approx 1.67 $, the system is underloaded (total offered load < number of servers), but still has some blocking due to finite capacity.

Would you like a Python code snippet to verify these calculations?