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A long line of ants find a hole in the pantry door. At time zero the first ant enters the pantry and then, one after the other, the ants pass through the hole at a rate of exactly one every minute. The ants progress through the hole is not random, however, suppose that the hole will remain open for a random time which follows an exponential distribution with an average time of 10 minutes. How many ants are expected to pass through the hole before it closes?

I don't really know how to begin this problem.

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  • $\begingroup$ What do you mean by "(the way) The ants progress through the hole is not random" ? $\endgroup$ – Jean Marie Dec 5 '19 at 23:21
  • $\begingroup$ I don't really understand this part too, by the way I just learned the expectation of sums of random variables, so I was trying to solve it by using that. $\endgroup$ – aswegsqw Dec 5 '19 at 23:31
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Assuming an ultra-strict interpretation of "the ants pass through the hole once a minute", we'll say an ant enters the hole at time $0, 1, 2$ and so on. Therefore if $T$ is exponentially distributed, we need to determine the expected value of $N=\lfloor T\rfloor + 1$ - where $\lfloor T\rfloor$ is the nearest integer to $T$, rounding down. Notice we need the $+1$ since the first ant enters at time $0$.

Working out the expected value of $\lfloor T\rfloor$ seems difficult at first, but in fact it's just a simple infinite sum. We have, by the definition of expected value:

$$E(\lfloor T\rfloor)=\sum_{n=0}^\infty nP(\lfloor T\rfloor=n)$$

and for every $n$, $P(\lfloor T\rfloor=n)=P(T\geq n, T<n+1)$.

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