Airline passengers arrive randomly and independently at the passenger-screening facility
at a major international airport. the mean arrival rate is 10 passengers per minute.
a. compute the probability of no arrivals in a one-minute period.
b. compute the probability that three or fewer passengers arrive in a one-minute period.
c. compute the probability of no arrivals in a 15-second period.
d. compute the probability of at least one arrival in a 15-second period.

Answer with Step-by-step explanation:

Since the random arrival process follows a poission's distribution

The probability of 'n' arrivals in time 't' is given by

where

is the average rate of arrival given as 10 passengers per minute

Part a)

Part b)

The probability that 3 or lesser passengers arrive in 1 minute is the sum of the probabilities of arrival of no passenger , 1 passenger , 2 passengers or 3 passengers respectively

Part c)

Since 15 seconds correspond to 0.25 minutes we have

Part d)

Probability of at least one arrival in 15 seconds is

ok

step-by-step explanation:

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4