Question

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.

a. Compute the probability of no arrivals in a one-minute period (to 6 decimals).

b. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals).

c. Compute the probability of no arrivals in a 15 second period (to 4 decimals).

d. Compute the probability of at least one arrival in a 15 second period (to 4 decimals).

Answer 1

The mean arrival rate here is given as 6 passengers per minute. Therefore the arrivals here could be modelled as:

a) The probability of no arrivals in a one-minute period is computed using the poisson distribution formula as:

Therefore 0.002479 is the required probability here.

b) The probability that three or fewer passengers arrive in a one-minute period is computed here as:

Therefore 0.1512 is the required probability here.

c) The probability that no arrivals arrive in a 15 second period that is in 15/60 = 0.25 minutes is computed here as:

= e^-0.25*6

= e^-1.5

= 0.2231

Therefore 0.2231 is the required probability here.

d) The probability that at least one arrival in a 15 sec time period happen is computed here as:

= 1 - Probability that no arrival happen in a 15 sec period

= 1 - 0.2231

= 0.7769

Therefore 0.7769 is the required probability here.