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So I have two independent random variables $X$ and $Y$, $Y$ ~ $U[0, 3]$ and the density function of $X$ is as follows:

$f(x) = 1/3$ if $0\le x \le 1$

$f(x) = 2/3$ if $1\le x \le 2$

$f(x) = 0$ otherwise

I calculated the joint density of this function by multiplying them, since they are independent. And now I'm trying to find $P(Y\gt X)$

Thanks in advance!

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  • $\begingroup$ if you think the answer is useful @bernas, can you please accept and/or upvote it? $\endgroup$ – gunes Feb 12 at 10:13
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The support is the rectangle $\mathcal{R}=[0,3]\times[0,2]$, and for $P(X>Y)$, you'll integrate the area under $Y=X$ line: $$\int_{\mathcal R} f_{X,Y}(x,y)dydx=\int_{0}^1\int_0^x \frac{1}{9} dy dx+\int_1^2\int_0^x\frac{2}{9} dydx$$

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  • $\begingroup$ Actually I made a mistake in the post, what I want is P(Y>X), what changes would have to be made? $\endgroup$ – bernas Feb 11 at 19:38
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    $\begingroup$ come on! subtract it from 1 $\endgroup$ – gunes Feb 11 at 19:38

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