Question 1200444: Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 11 passengers per minute.
a. Compute the probability of no arrivals in a one-minute period (to 6 decimals).
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! **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.**
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