Question

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 9 passengers per minute.
Compute the probability of no arrivals in a one-minute period (to 6 decimals).
Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals).
Compute the probability of no arrivals in a 15-second period (to 4 decimals).
Compute the probability of at least one arrival in a 15-second period (to 4 decimals).

Answer 1

A) probability of no arrivals in a one-minute period will be;
f(0)=[100×e−10]0!
f(0)=[100×e−10]0!
f(0) = 0.00004539993
To six decimal places gives;
f(0) = 0.000045

B) the probability that three or fewer passengers arrive in a one-minute period is given by;
P(x≤3)=f(0)+f(1)+f(2)+f(3)
P(x≤3)=f(0)+f(1)+f(2)+f(3)
We already have f(0) = 0.000045
f(1)=[101×e−10]1!
f(1)=[101×e−10]1!
f(1) = 0.000454

f(2)=[102×e−10]2!
f(2)=[102×e−10]2!
f(2) = 0.00227

f(3)=[103×e−10]3!
f(3)=[103×e−10]3!f(3) = 0.007567

Thus;
P(x≤3)=0.000045+0.00227+0.007567=0.009882
P(x≤3)=0.000045+0.00227+0.007567=0.009882
To four decimal places gives;
P(x≤3)=0.0099
P(x≤3)=0.0099

C) no arrivals in a 15-second period means that;
μ=10×1560μ=10×1560μ=2.5μ=2.5
Thus;
the probability of no arrivals in a 15-second period ia;
f(0)=[2.50×e−2.5]0! f(0)=[2.50×e−2.5]0!
f(0) = 0.082085
To four decimal places is;
f(0) = 0.0821

D) the probability of at least one arrival in a 15-second period will be gotten from the complement rule. Thus;
P(x≥1)=1−f(0)P(x≥1)=1−f(0)P(x≥1)=1−0.0821P(x≥1)=1−0.0821P(x≥1)=0.9179P(x≥1)=0.9179