POISSON (Random) EVENTS

2

What is the probability of 3 or fewer vehicles arriving in 1 minute if the arrival rate is 120 vph?

II.

P(n <= N) = Sum from 0 to N for: P(n)

Remember:

P(n <= 3) = Sum from 0 to 3 for: P(n)

X0 = 1

P(n <= 3) = P(0) + P(1) + P(2) + P(3)

0! = 1

P(0)=e-(120veh/hour)(1min*1 hr./60 min)[(120veh/hr)(1min*1hr./60min)]0/0! = 0.135

P(1)=e-(120veh/hour)(1min*1 hr./60 min)[(120veh/hr)(1min*1hr./60min)]1/1! = 0.271

P(2)=e-(120veh/hour)(1min*1 hr./60 min)[(120veh/hr)(1min*1hr./60min)]2/(1*2) = 0.271

P(3)=e-(120veh/hour)(1min*1 hr./60 min)[(120veh/hr)(1min*1hr./60min)]3/(1*2*3) = 0.180

P(n <= 3) = 0.135 + 0.271 + 0.271 + 0.180 = 0.857

What is the probability of more than 3 vehicles arriving in 1 minute if the arrival rate is 120 vph?

III.

P(n > N) = 1 - P(n <= N)

Remember:

P(n > 3) = 1 - P(n <= 3)

X0 = 1

P(n > 3) = 1 - 0.857 = 0.143

0! = 1