Airline passengers arrive randomly and independetly at the passenger screening facility at a major international airport. the mean arrival rate is 10 passengers per minute. compute the probability of no arrivals in a 15 second period. Solution Mean in 15 sec period is 10/4= 2.5. So using the Poisson distribution the probability of no arrivals is e^(-2.5)= 0.08208.

1. Airline passengers arrive randomly and independetly at the passenger screening facility at a
major international airport. the mean arrival rate is 10 passengers per minute. compute the
probability of no arrivals in a 15 second period.
Solution
Mean in 15 sec period is 10/4= 2.5. So using the Poisson distribution the probability
of no arrivals is e^(-2.5)= 0.08208