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Suppose $U$ and $V$ are independent, standard uniforms random variables. Suppose I want to calculate the probability $P(U^2 > 4V)$. I would like to use some intelligent method like starting from a joint density of (U, V), then transform the problem into a simpler one, to be solved using some geometry on the square $[0,1] \times [0,1]$. Is it possible? Do you know some smart solution to this problem? Thanks in advance.

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Since $U$ and $V$ are independent, the joint density is a constant in the square of $(u,v) \in [0,1]\times[0,1]$. Therefore $P(U^2 > 4V)$ equals to the area of $v < u^2/4$ within the square. $$P(U^2 > 4V) = \int_0^1\frac{u^2}4\,du = \frac1{12}.$$

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