Given its joint probability density function, find $P\left\{Y>\frac{1}{2}|X<\frac{1}{2}\right\}$.

The joint probability density function of $X$ and $Y$ is given by

$f(x,y)=\frac{6}{7}\left(x^{2}+\frac{xy}{2}\right)$, $0<x<1,0<y<2$.

Find $P\left\{Y>\frac{1}{2}|X<\frac{1}{2}\right\}$. (conditional probability)

My notes only have the case where, for example, given $Y=y$, find $P\left\{X<a|Y=y\right\}$, which has the formula $P\left\{X<a|Y=y\right\}=\int_{-\infty}^{a}f_{X|Y}(x|y)dx$, where $f_{X|Y}(x|y)=\frac{f(x,y)}{f_{Y}(y)}$.

Any idea how do I do the case where the condition isn't constant?

Solution 1:

$$P(y>\frac{1}{2}|X<\frac{1}{2})=\frac{P(Y>\frac{1}{2},X<\frac{1}{2})}{P(X<\frac{1}{2})}$$.

$$=\frac{\int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}}^{2}f(x,y)\,dydx}{\int_{0}^{\frac{1}{2}}\int_{0}^{2}f(x,y)\,dydx}$$