# Poisson distribution, mean time and probability of waiting

We know that average number of planes landing at particular airport during an hour is 36.

a) what is the mean time for waiting for the first landing during an hour?

b) find probability of waiting more than 1/2 hour to see the first landing?

I assumed this is Poisson distribution as Exponential is memoryless (each minute would have independent probability, am I right?)

But to be honest I don't really know what to do next. I wrote down the data from the task:

$$\lambda = 36$$

And I'm stuck. Looking more for tips how to solve this task, not full solution.

Thank you for help in advance.

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For part (a), you are given that on average $36$ planes are observed every hour. In this case, on average, how many hours does it take to observe a plane? What is this in minutes?

For (b), if $36$ planes land in an hour, how many planes land in half an hour on average? Then what is the probability of seeing no planes in the first half hour?

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so a) 36/60, sooooo 0,6 minutes?, is the time, right? –  mickula Jan 21 '13 at 20:16
Not quite. You would expect to see a plane every $1/36$ hours, or $60/36=1\frac{2}{3}$ minutes –  Daniel Littlewood Jan 21 '13 at 20:36
Well, point for you. Thanks for explaining. When it comes to b) I have to do that foreach k := 1 to 30 probabilities are $0$ and... what next? $P(X < 30 ) = 1-e^{- \lambda k} + e^{- \lambda k+1}$ etc ? –  mickula Jan 21 '13 at 21:03
No problem :) The number of planes you see in a half hour has the distribution $Po(18)$ (since it is half the time). Then $P(X=0)=e^{-18}$. –  Daniel Littlewood Jan 22 '13 at 17:20

Hint: The Poisson with $\lambda=36$ is indeed the intended model. Then the waiting time is exponential, parameter $36$. What is the probability that an exponential with parameter $\lambda=36$ is greater than $\frac{1}{2}$? It can be expressed as an integral, though you probably already know an explicit formula for the cdf of an exponential, so will not need to integrate.

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$P(X < \frac{1}{2} ) = 1-e^{- \lambda \frac{1}{2}}$ ? –  mickula Jan 21 '13 at 20:18
Yes. The expression for $\Pr(X\gt 1/2)$ is even simpler, $e^{-(\lambda)(1/2)}=e^{-18}$. Here we used relationship between Poisson and exponential. One can also treat it in purely Poisson terms, number $W$ of landings in time $1/2$ is Poisson parameter $(36)(1/2)$. –  André Nicolas Jan 21 '13 at 21:43