Question 1200444
**1. Identify the Distribution**

* Since passengers arrive randomly and independently at a constant average rate, this situation can be modeled by a **Poisson distribution**.

**2. Poisson Probability Formula**

* The probability of observing *k* events in a given time interval, given the average rate (λ) of events occurring in that interval, is:

  * P(X = k) = (λ^k * e^(-λ)) / k! 

  where:
    * X is the number of events (passengers)
    * λ is the average arrival rate (11 passengers/minute)
    * k is the desired number of events (0 passengers in this case)
    * e is the base of the natural logarithm (approximately 2.71828)
    * k! is the factorial of k (0! = 1)

**3. Calculate Probability of No Arrivals**

* P(X = 0) = (11^0 * e^(-11)) / 0! 
* P(X = 0) = (1 * e^(-11)) / 1
* P(X = 0) = e^(-11) 
* P(X = 0) ≈ 0.0000000016 

**Therefore, the probability of no arrivals in a one-minute period is approximately 0.0000000016.**