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Problem.1) The joint probability density function of two continuous random variables $X$ and $Y$ is given by

$$ f(x) = \begin{cases} \frac{1}{8}(x+y), & \text{for } 0 \leq x \leq 2, 0 \leq y \leq 2\\ 0, & \text{otherwise} \end{cases} $$

Calculate $\mathbf{P}(X + Y \leq 2)$.


1) The original image can be seen in https://i.stack.imgur.com/Xbmhq.jpg

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  • $\begingroup$ thank you all, I am wtill not good at this. $\endgroup$ – Sergio Jun 30 at 11:08
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$P(X+Y\leq 2)=\int_0^{2}\int_0^{2-x} \frac 1 8 (x+y)dydx=\frac 1 8 \int_0^{2} [x(2-x)+\frac {(2-x)^{2}} 2] dx$. Can you continue?

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  • $\begingroup$ thank you! all clear now. $\endgroup$ – Sergio Jun 30 at 18:02

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