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 May 26th, 2013, 02:47 AM #1 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 probability of X and Y The joint propability function of X and Y is given by $f(x,y)= \left\{\begin{matrix} 2 & x> 0,y> 0,x+y< 1\\ 0 & elswhere \end{matrix}\right$. So far I found $P(X\leq \frac{1}{2},Y\leq \frac{1}{2})$. I still need to find $P(X+Y> \frac{2}{3})$ and $P(X> 2Y)$. The answer should be $\frac{5}{9}$ and $\frac{1}{3}$, respectively. But I do not see how to get there. For the first I take the double integral over 2 with $0< x< 1$ and $\frac{2}{3}-x< y< 1-x$. For the second I take boundaries $0< y< 1$ and $2y< x< 1-y$. The answer I get are 1 and 0. Why is this wrong and how should it be done?
 May 26th, 2013, 03:02 AM #2 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 Re: probability of X and Y Now I took $1-P(X+Y< \frac{2}{3})$. I made a graph of $P(X+Y< \frac{2}{3})$ and took the boundaries $0< x< \frac{2}{3}$ and $0< y< \frac{2}{3}$. That gave me $\frac{4}{9}$, so the first one is solved
 May 26th, 2013, 03:13 AM #3 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 Re: probability of X and Y And a sketch also did the work for the other question. I took $0 and $2y and that gave me indeed the right answer. Sorry for bothering and thanks anyway!
May 26th, 2013, 03:46 AM   #4
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Re: probability of X and Y

[color=#000000]It is much easier to make the probability computations using area computation.

[attachment=0:m6xn6toy]george.png[/attachment:m6xn6toy]

$\mathbb{P}\left(\mathbb{X}=&\mathbb{Y}\leq \frac{1}{2}\right)=2\cdot \text{area of green square}=2\cdot\frac{1}{4}=\frac{1}{2}$

$\mathbb{P}\left(\mathbb{Y}=>\frac{2}{3}-\mathbb{X}\right)=1-\mathbb{P}\left(\mathbb{Y}\leq \frac{2}{3}-\mathbb{X}\right)=1-2\cdot \text{area of red triangle}=1-2\cdot \frac{2}{9}=\frac{5}{9}$

$\mathbb{P}\left(\mathbb{Y}=<\frac{\mathbb{X}}{2}\ri ght)=2\cdot\text{area of cyan triangle}=2\cdot\frac{1}{6}=\frac{1}{3}$[/color]
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 May 26th, 2013, 04:03 AM #5 Member   Joined: Apr 2013 Posts: 42 Thanks: 1 Re: probability of X and Y The graph indeed makes it perfectly clear. I'll remember the tip for other questions/problems. Thanks a lot!

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