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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. – Did Dec 18 '12 at 12:24
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. – Noah Snyder Dec 18 '12 at 12:26
@prasenjit On the other hand, I would like to know some context behind this question. – dtldarek Dec 18 '12 at 13:27
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). – dtldarek Dec 18 '12 at 13:54
@prasenjit: Your question still lacks a hypothesis, which is that, presumably, X and Y are independent. – Did Dec 18 '12 at 14:58

Well although I doesn't make sense to put the words prove and probability in the same sentence, I'll answer this anyway. The question should be rephrased as find the probability and not prove the probability.

Anyway, the simplest definition of probability is No. of favorable outcomes divided by Total NO. of outcomes. Since, we are selecting a point from the square, the area of the square is the total no of outcomes. And, the favorable is the area of the circle, since, that is where we want the point. So, answer is area of circle divided by area of square.

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Uh. Your first sentence is just plain false. Was it meant as a joke? – cardinal Dec 18 '12 at 13:02
Where do you see a number of favorable outcomes in the area of a disk? – Did Dec 18 '12 at 13:29

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