Question

On Fridays at KK airport in Lusaka, airplanes arrive at an average of 3 for the one hour period 13 00 hours to 14 00 hours. If these arrivals are distributed according to the Poisson probability distribution, what are the probability that either one or two airplanes will arrive between 13.00 hours and 14 00 hours next Friday?   

Answer 1

Here given Average = 3

Probability of x successes in poisson distribution = (^x * e^-) / x!

Value of e = 2.7183

We need to find P(X=1) + P(X=2) here since we need to find probability that either one or two airplanes will arrive between 13.00 hours and 14 00 hours next Friday

P(X=1) + P(X=2) = (3¹ * 2.7183^-3) / 1! + (3² * 2.7183^-3) / 2!

= 0.149361 + 0.224042

= 0.373403

So Probability that either one or two airplanes will arrive between 13.00 hours and 14 00 hours next Friday = 0.373403