Question

Airline passengers arrive randomly and independently at the passenger-screening facility

Answer 1

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

$P(n,t)=\frac{(\lambda t)^ne^{-\lambda t}}{n!}$

where

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

Part a)

$P(0,1)=\frac{(10)^0\cdot e^{-10}}{0!}=0.0000454$

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

$P(E)=\frac{(10)^0e^{-10}}{0!}+\frac{(10)^1e^{-10}}{1!} +\frac{(10)^2e^{-10}}{2!}+\frac{(10)^3e^{-10}}{3!}=0.01033$

Part c)

Since 15 seconds correspond to 0.25 minutes we have

$P(0,0.25)=\frac{(10\times 0.25)^0\cdot e^{-10\times 0.25}}{0!}=0.082$

Part d)

Probability of at least one arrival in 15 seconds is

$P=(1-p(0,0.25)\\\\P=1-0.082=0.918$