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For the given joint pdf of X and Y $$f(x,y) = 8xy, 0≤ y≤ x≤ 1$$

I need to find $$P(X<3/4 | Y<1/2)$$ I'm not quite sure whether my approach is right, but here's what I did. I found the marginal pdf of Y, which is $$4y - 4y^3, 0≤ y≤ 1$$

I know that $$P(X<3/4 | Y<1/2) = P(X<3/4,Y<1/2) / P(Y<1/2)$$

Now I have $$P(X<3/4,Y<1/2) = \int_{0.5}^{0.75}\int_{0}^{0.5} (8xy) dy dx $$ What I am unsure about are the limit values. Are they correct? And does it matter whether I integrate with respect to y or x first?

Following that, I get $$P(X<3/4,Y<1/2) = 5/32$$ Then I calculate $$P(Y<1/2) = \int_{0}^{0.5} (4y - 4y^3) dy = 7/16$$ Thus$$P(X<3/4 | Y<1/2) = (5/32) / (7/16) = 5/14$$

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By definition of conditional probability, $$ \mathbb P\left(X<\frac34\mid Y<\frac 12\right) = \frac{\mathbb P(X<\frac34,Y<\frac12)}{\mathbb P(Y<\frac12)}. $$ We compute the marginal density of $Y$ by integrating over the range of values for $X$ - for $0<y<1$ we have \begin{align} f_Y(y) &= \int_y^1 f_{X,Y}(x,y)\ \mathsf dx\\ &= \int_y^1 8xy\ \mathsf dx\\ &= 4y(1-y^2). \end{align} It follows readily that $$ \mathbb P\left(Y<\frac12\right) = \int_0^{\frac12} f_Y(y)\ \mathsf dy= \int_0^{\frac12}4y(1-y^2) = \frac7{16}. $$ Now write $$ \left\{X<\frac34,Y<\frac12\right\} = \left\{X<\frac12,Y<\frac12\right\} \cup \left\{\frac12<X<\frac34,Y<\frac12\right\}. $$ We compute $$ \mathbb P\left(X<\frac12,Y<\frac12\right) = \int_0^{\frac12}\int_y^{\frac12} 8xy\ \mathsf dx \ \mathsf dy = \frac1{16} $$ and $$ \mathbb P\left(\frac12<X<\frac34,Y<\frac12\right) = \int_0^{\frac12}\int_{\frac12}^{\frac34} 8xy\ \mathsf dx \ \mathsf dy = \frac5{32}. $$ This yields \begin{align} \mathbb P\left(X<\frac34\mid Y<\frac 12\right) &= \frac{\mathbb P\left(X<\frac12,Y<\frac12\right) + \mathbb P\left(\frac12<X<\frac34,Y<\frac12\right)}{\mathbb P\left(Y<\frac12\right)} \\ &= \frac{\frac1{16}+\frac5{32}}{\frac7{16}}\\ &= \frac12. \end{align}

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