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Behind a rectangle grid evenly (i.e. uniform distribution) scattered dots. Could it be considered identical (will have the same uniform distribution) to a sequence of independent events with probability $\frac{1}{N}$ to hit the cell with dot? (Where $N$ is a total number of cells of the grid: $N_1 \times N_2$) And if it so, is it right to say that the way you choose cells on the grid doesn't matter and the probability to find the dot would be the same as $\frac{1}{N}$.

How to prove it (or disprove), what the assertions should be involved? And how to extend the proof on 3-dimensional space and further on $n$-dimensions?

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  • $\begingroup$ What does $N$ mean? $\endgroup$ – Gerry Myerson Aug 17 '13 at 23:52
  • $\begingroup$ Please edit that information into the body of the question. People shouldn't have to go through the comments to know what the question is asking. $\endgroup$ – Gerry Myerson Aug 17 '13 at 23:57
  • $\begingroup$ Some aspects of the question are still unclear to me. But initially I'd presume that "can it be considered the same?" means "Is the probability distribution the same?". $\endgroup$ – Michael Hardy Aug 18 '13 at 0:10
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If I understand correctly you are telling me you have an $M\times M$ square where $M\times M=N$. Then you are placing the dot at some random location in the $M\times M$ square. If each location has the same probability (uniformly distributed) what are the chances of finding the dot? Then you would be correct in saying you have a $1/N$ chance of finding the dot and the $M\times M$ space is equivalent to a line of length $N$. In general if you have a sample space of size $N$ and each event is equally likely to occur then your chance of the even occurring (in this case choosing your dot) is just $1/N$.

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  • $\begingroup$ I feel the same, but I want to have mathematical basis and rigor. $\endgroup$ – rook Aug 18 '13 at 8:52

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