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. What is the probability of no arrivals in a 1-minute period? If required, round your answer to six decimal places. .000123 What is the probability of 3 or fewer arrivals in a 1-minute period? If required, round your answer to six decimal places. .021226 What is the probability of no arrivals in a 16-second period? If required, round your answer to four decimal places. .0907 What is the probability of at least 1 arrival in a 16-second period? If required, round your answer to four decimal places. .3084

Answer 1

Answer with Step-by-step explanation:

Since the random arrival of passengers follow Poission's distribution

The probability that 'n' passengers arrive in 't' time with average arrival rate 'q' is given by

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

Given q  = 9 passengers per minute

Part a)

The probability that no passenger arrives in 1 minute equals

$P(0,1)=\frac{(9\times 1)^0e^{9\times 1}}{0!}\\\\P(E)=0.000123$

Part b)

The probability of lesser than 3 arrivals occur equals the sum of probabilities of

1) No arrival

2) Exactly 1 arrival

3) Exactly 2 arrivals

4) Exactly 3 arrivals

$P(E)=p(0,1)+p(1,1)+p(2,1)+p(3,1)=0.000123+0.0011+0.004998+0.014994\\\\\therefore P(E)=0.212256$

Part c)

Probability of at least 1 arrival in 16 seconds euals

P(1,16/60)

$P(1,16/60)=\frac{(9\times \frac{16}{60})^1e^{-9\times \frac{16}{60}}}{1!}\\\\\therefore P(1,16/60)=0.2177$