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The random variables X and Y are jointly distributed according to the pdf $f_{X,Y} (x,y)=x+y, 0 <x<1, 0<y<1$. Find $P(Y>1/3|X=1/2)$

Since this is a conditional probability, can't I just write $P(Y>1/3|X=1/2)=P(Y>1/3 , X=1/2)*P(X=1/2)=0$ since $P(X=1/2)=0$ -because the random variables are continuous??

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I apologize in advance if this problem was that easy, it is for my practice final and I'm used to doing double integration with these Joint PDFs. – The Substitute May 12 '12 at 21:24
You have your conditional equation wrong. You might have tried $\Pr(Y \gt \frac13 | X = \frac12)\times \Pr(X=\frac12) = P(Y \gt \frac13 , X=\frac12)$ but that leads towards $\frac00$ for the conditional probability. – Henry May 12 '12 at 22:31
up vote 1 down vote accepted

If $X=\frac{1}{2}$ then the conditional density for $Y$ is proportional to $\frac{1}{2}+y$, so $$\Pr(Y \gt \tfrac{1}{3}|X=\tfrac{1}{2}) = \frac{\int_{y=\frac{1}{3}}^1 (\frac{1}{2}+y)dy}{\int_{y=0}^1 (\frac{1}{2}+y)dy} = \frac{7}{9}.$$

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