Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 6 passengers per minute. Compute the probability of no arrivals in a one-minute period (to 6 decimals).

The probability of no arrivals in a one-minute period is 0.002479.

Step-by-step explanation:

We are given that Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 6 passengers per minute.

The above situation can be represented through Poisson distribution because a random variable which includes the arrival rate is considered as Poisson random variable.

The Probability distribution function for Poisson random variable is ;

where,   = arrival rate

Let X = Arrival of Airline passengers

The mean of the Poisson distribution is given by = E(X) = , which is given to us as 6 passengers per minute.

So, X ~ Poisson ()

Now, the probability of no arrivals in a one-minute period is given by = P(X = 0)

      P(X = 0) =  

                    =

                    = 0.002479

Hence, the probability of no arrivals in a one-minute period is 0.002479.

0.002478 is the probability of no arrivals in a one-minute period.  

Step-by-step explanation:

We are given the following information in the question:

Mean arrival rate = 6 passengers per minute.

The airline passengers can be treated as a Poisson distribution.

Poisson distribution.

Formula:

P( no arrivals in a one-minute period)

0.002478 is the probability of no arrivals in a one-minute period.

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step-by-step explanation: