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Suppose a farmer shoots at $n$ targets. He hits each with iid probability $p$. Of course, he hits $np$ in expectation. The same targets remain in the same position, and a new farmer comes and shots. He hits again with probability $p$. After he shoots, farmers keep coming and shooting with the same probability $p$ of hitting the same $n$ targets. Suppose $X$ farmers came. How many targets have been hit (at least once)?

If the targets were different, the answer would be trivially $Xnp$. But I am having problems to account for the fact that they are the same targets.

fox
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  • Do you want the expected number? (This is what your last paragraph suggests.) – Clement C. May 05 '18 at 18:55
  • You can think of it this way: since each farmer's shot at each target is independent of any other farmer's shots at any target, let all the farmers take shots at the first target, then all farmer take shots at the second target, etc. Then you easily get results like the answers below. – David K May 05 '18 at 19:04

2 Answers2

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The probability that a fixed target is not hit by any of the $X$ farmers is $ (1-p)^X $, since all farmers shoot independently. Thus, the probability that this fixed target is hit at least once is $$ 1-(1-p)^X\,. $$ By linearity of expectation, the expected number of targets hit at least once is $$ \mathbb{E}\left[\sum_{i=1}^n \mathbb{1}_{i \text{ hit at least once}}\right] = \sum_{i=1}^n \mathbb{E}[\mathbb{1}_{i \text{ hit at least once}}] = \sum_{i=1}^n \mathbb{P}\{i \text{ hit at least once}\} = \boxed{n (1-(1-p)^X)}\,. $$ As a sanity check, for $X=1$ you indeed retrieve the value $np$.

Clement C.
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For $i=1,\dots,n$ let $T_i$ takes value $1$ is target $i$ is hit by one of the farmers and let it take value $0$ otherwise.

Then $T=\sum_{i=1}^nT_i$ targets are hit in total and with linearity of expectation and symmetry we find:$$\mathsf ET=\sum_{i=1}^n\mathsf ET_i=n\mathsf ET_1=n\mathsf P(T_1=1)=n(1-\mathsf P(T_1=0))=n(1-(1-p)^X)$$

drhab
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  • Thank you both. I was asking as a way to solve this more general problem, maybe you have some idea about it? https://math.stackexchange.com/questions/2768095/how-many-paths-of-length-2-in-a-general-random-graph?noredirect=1#comment5708710_2768095 – fox May 05 '18 at 20:01