# How many ants are expected to pass through the hole before it closes?

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.

• What do you mean by "(the way) The ants progress through the hole is not random" ? – Jean Marie Dec 5 '19 at 23:21
• 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. – aswegsqw Dec 5 '19 at 23:31

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.