Find $\operatorname{Pr}\left ( Y< 0.5\mid X< 0.5 \right )$ with the given joint probability density function

Question

Given two random variables $X$ and $Y.$ Find $\operatorname{Pr}\left ( Y< 0.5\mid X< 0.5 \right )$ with the joint density function $$f\left ( x, y \right )= \left\{\begin{matrix} 6xy & {\rm if}\,0\leq x\leq 1, 0\leq y\leq 2- 2x\\ 0 & {\rm otherwise} \end{matrix}\right.$$ I think I need to find the marginal density functions of $X$ and of $Y$ but I don't know how.
If successful, then we use double integrals to find $\operatorname{Pr}\left ( Y< 0.5, X< 0.5 \right )$ and $\operatorname{Pr}\left ( X< 0.5 \right ).$
I will continue the @tommik's idea as following. We obtain $$\operatorname{Pr}\left ( Y< 0.5, X< 0.5 \right )= \frac{\int_{0}^{0.5}\int_{0}^{0.5}6xy{\rm d}x{\rm d}y}{\int_{0}^{1}\int_{0}^{2}6xy{\rm d}x{\rm d}y}= \frac{\left ( 1/8 \right )^{2}}{1/2\cdot 2}= \frac{1}{64}$$ $$\operatorname{Pr}\left ( X< 0.5 \right )= \frac{\int_{0}^{0.5}12x\left ( 1- x \right )^{2}{\rm }dx}{\int_{0}^{1}12x\left ( 1- x \right )^{2}{\rm d}x}= \frac{11/192}{1/12}= \frac{11}{16}$$ $$\Rightarrow\operatorname{Pr}\left ( Y< 0.5\mid X< 0.5 \right )= \frac{1/64}{11/16}= \frac{1}{44}$$

Answer 1

Simply use the definition

$$P(Y<0.5|X<0.5)=\frac{P(X<0.5,Y<0.5)}{P(X<0.5)}$$

The marginal density of X is the following

$$f_X(x)=12x(1-x)^2$$

thus...