Air traffic control stations often have insufficient numbers of air traffic controllers, sometimes just one person on duty. In a recent study, a lone air traffic controller is managing an airstrip in which the expected time between arrivals of airplanes is $15$ minutes. Assume that the times between consecutive arrivals are independent Exponential random variables. If he falls asleep for a period of $5$ minutes, what is the probability that one or more airplanes arrive during this period (but cannot land because the air traffic controller is not awake to guide them)?
$E(X) = 15$ minutes where X is exponential and describes one plane. so $\lambda = 1/15$ per minute. So $\lambda = 1/3$ every $5$ minutes
So let $Y$ be the number of planes that land while the air traffic ccontroller is asleep.
I believe that $Y$ is poisson since it's a fixed time interval that the air traffic controller is asleep.
So I think the answer is $P(Y\geq 1) = 1 - P(Y=0) = 1 - \cfrac{e^{-1/3}\cdot\cfrac{1}{3}^0}{0!} \approx .2835$. So the probability that one or more airplanes arrive during the $5$ minute period he's asleep is $.2835$
Do I have this right? If not what I am doing wrong?
Yes, that is correct.
Another line of reasoning is to observe that, given the controller has fallen asleep at time $t = 0$, what is the probability that the next plane to arrive does so within the next $5$ minutes; i.e., if $T$ is the next time to arrival (in minutes) random variable, what is $\Pr[T \le 5]$? Note $T \sim \operatorname{Exponential}(\lambda = 1/15)$, because planes have arrival intensity of $15$ per minute. So $$\Pr[T \le 5] = 1 - e^{-5 \lambda} = 1 - e^{-5/15} = 1 - e^{-1/3},$$ same as what you wrote.