Conditional probability with two inequalities

Question

Let $(X,Y)$ a random vector with density $f(x,y)=\frac{6}{7}(x^2+\frac{xy}{2});0<x<1,0<y<2$.

After checking that is a density, finding the marginal $f_X(x)=\frac{12}{7}x^2+\frac{6}{7}x$ and the probability $\mathbb{P}(X>Y)=\frac{15}{56}$, i have to find $\mathbb{P}(Y>\frac{1}{2}|X<\frac{1}{2})$.

I know that the domain is $0<x<\frac{1}{2}<y<2$, so i applied the conditional probability $\mathbb{P}(Y>\frac{1}{2}|X<\frac{1}{2})=\frac{\mathbb{P}((Y>\frac{1}{2})\cap (X<\frac{1}{2}))}{\mathbb{P}(X<\frac{1}{2})}$ and i tried to integrate in the most different ways…

1) $\frac{\int_{0}^{3/2}(\int_{0}^{5/2}f(x,y)dy)dx}{\int_{0}^{1/2}f_X(x)dx}$

2) $\frac{\int_{0}^{1/2}(\int_{1/2}^{2}f(x,y)dy)dx}{\int_{0}^{1/2}f_X(x)dx}$

3) ${\int_{0}^{1/2}(\int_{1/2}^{2}\frac{f(x,y)}{f_X(x)}dy)}dx$

…but to no avail.

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How to act when the conditional probability provides for two disequalities?

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EDIT: Trivially, you might say "Making the graph". Well, I'm having a hard time to understand how to draw it correctly. In the first probability that i said, the graph was fairly simple because I could use the 1°-3° bisector to separate the region to integrate.

Answer 1

You found correctly: $$\int_{0}^1\int_0^2f(x,y)dydx=\int_{0}^1\int_0^2\frac{6}{7}(x^2+\frac{xy}{2}) dydx=1\\ f_X(x)=\int_0^2 \frac{6}{7}(x^2+\frac{xy}{2})dy=\frac{12}{7}x^2+\frac{6}{7}x.$$ Now, your second formula is correct: $$\mathbb{P}(Y>\frac{1}{2}|X<\frac{1}{2})=\frac{\mathbb{P}((Y>\frac{1}{2})\cap (X<\frac{1}{2}))}{\mathbb{P}(X<\frac{1}{2})}=\\ \frac{\int_{0}^{1/2}(\int_{1/2}^{2}f(x,y)dy)dx}{\int_{0}^{1/2}f_X(x)dx}=\\ \frac{\int_{0}^{1/2}\frac67(x^2y+\frac{xy^2}{4})|_{1/2}^2dx}{\int_{0}^{1/2}(\frac{12}{7}x^2+\frac{6}{7}x)dx}=\\ \frac{\int_{0}^{1/2}(\frac97x^2+\frac{45x}{112})dx}{(\frac{4x^3}{7}+\frac{3x^2}{7})|_{0}^{1/2}}=\\ \frac{(\frac{3x^3}{7}+\frac{45x^2}{224})|_{0}^{1/2}}{5/28}=\\ \frac{69/448}{5/28}=\\ \frac{69}{80}\approx 0.8625.$$

Answer 2

For $a\in\left[0,1\right]$ and $b\in\left[0,2\right]$ define:

$$\begin{aligned}f\left(a,b\right) & :=P\left(X<a,Y>b,\right)\\ & =\frac{6}{7}\int_{0}^{a}\int_{b}^{2}x^{2}+\frac{1}{2}xy\;dydx\\ & =\frac{6}{7}\int_{0}^{a}\left[x^{2}y+\frac{1}{4}xy^{2}\right]_{b}^{2}\;dx\\ & =\frac{6}{7}\int_{0}^{a}\left(2-b\right)x^{2}+\frac{1}{4}\left(4-b^{2}\right)x\;dx\\ & =\frac{6}{7}\left[\frac{1}{3}\left(2-b\right)x^{3}+\frac{1}{8}\left(4-b^{2}\right)x^{2}\right]_{0}^{a}\\ & =\frac{6}{7}\left(\frac{1}{3}\left(2-b\right)a^{3}+\frac{1}{8}\left(4-b^{2}\right)a^{2}\right)\\ & =\frac{1}{28}a^{2}\left(2-b\right)\left(8a+6+3b\right) \end{aligned} $$

Then $P\left(X<\frac{1}{2}\right)=f\left(\frac{1}{2},0\right)$ and $P\left(Y>\frac{1}{2},X<\frac{1}{2}\right)=f\left(\frac{1}{2},\frac{1}{2}\right)$ so that: $$P\left(Y>\frac{1}{2}\mid X<\frac{1}{2}\right)=\frac{f\left(\frac{1}{2},\frac{1}{2}\right)}{f\left(\frac{1}{2},0\right)}$$

Just substitute.