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How to prove that the probability that a randomly selected point in a square lies inside the circle inscribed in the square is equal to the ratio of the area of the circle and square ?

To rephrase the question in a better sense, suppose $X$ and $Y$ be the independent random variables representing the $x$ and $y$ co-ordinates of the point. Here $X$ and $Y$ are uniformly distributed between $[-a,a]$ where the square has vertices with co-ordinates $(\pm a, \pm a)$. Now I want to find the probability $P(X^{2}+Y^{2} < a^{2})$. Hope I have made my question clear.

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    You cannot prove it, this is (or is not) a hypothesis.2012-12-18
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    When you say "random" here without modifying it, what you mean precisely is that the odds of it lying in some region is exactly the ratio of the area of that region to the area of the whole thing.2012-12-18
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    @prasenjit On the other hand, I would like to know some context behind this question.2012-12-18
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    @dtldarek Would you be presuming that I downvoted, by any chance? I did not (and I am a little fed up with this kind of assumption, please do not assign your ways to me).2012-12-18
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    I don't know why this question was downvoted. It might be incomplete, but surely there were worse which were treated far better way. The main question is, what is the probability space (and that we can guess). Once we know the space, we can prove statements such as $P(A)= \frac{π}{4}$ (e.g. by calculating $P(A)$, but it is still a proof).2012-12-18
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    @did I didn't and it was just after your comment that I realized what I wrote could be understood this way. The notification was due to the part "you cannot prove it". Fixed now. But you could elaborate more on "this is a hypothesis"; prove or disprove $P(A) = \frac{\pi}{4}$ seems valid enough for me, why do you object?2012-12-18
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    @prasenjit: Your question still lacks a hypothesis, which is that, presumably, X and Y are independent.2012-12-18
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    @dtldarek Because to prove that, once the necessary hypothesis are correctly stated, is one tenth of the exercise (the other nine tenths being to state correctly the hypothesis, something which is almost done at present, but not quite). (Nota: You were right to delete your previous comment.)2012-12-18
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    So, how to find this probability ?2012-12-18
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    Legitimate question. This is the basis of increasingly important Monte Carlo algorithms in which points are generated pseudo-randomly, eg: http://physics.gac.edu/~huber/envision/instruct/montecar.htm. Clearly, in addition to the independence and uniform distribution of samples, you would need an infinite sequence of points to converge to the ratios of the areas of circle/square. Central limit.2012-12-18
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    @alancalvitti: Can you elaborate your solution ?2012-12-19
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    @prasenjit, I started developing a Mathematica notebook for this, but then realized it is described here: http://en.wikipedia.org/wiki/Monte_Carlo_integration2012-12-28
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    @alancalvitti, I went through this, but could understand the proof of $ \lim_{N \to \infty} Q_N = I $.2013-01-12
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    @alancalvitti: *could not. My question is how to prove the correctness of Monte Carlo integration technique.2013-01-13
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    @prasenjit, let me think about this - it's not my area of expertise but it should be simple using law of large numbers (probably weak form) and the knowledge that independent samples means that the the joint distribution factorized completely into the product of the individual uniforly distributed random variables: $P(x_1,...x_n)=P(x_1)*...*P(x_n)$2013-01-13

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