The joint probability density function of $X$ and $Y$ is $f(x,y)=3y$ for $0\le x\le y\le 1$ and 0 for otherwise. Find $$P \left(X<\frac{2}{3}\Bigg| Y=\frac{1}{3} \right)$$
I am confused about this one since $$ P \left(X<\frac{2}{3}\Bigg|Y=\frac{1}{3} \right)=\frac{P \left(X<\frac{2}{3},Y=\frac{1}{3} \right)}{P \left(Y=\frac{1}{3} \right)} $$
I can get the marginal of $Y$: $$ f_Y(y)=\int_{0}^y f(x,y) \, dx=3y^2, 0\le y\le 1. $$
So $P \left(Y=\frac{1}{3} \right)=1/3$. But what is $P \left(X<\frac{2}{3},Y=\frac{1}{3} \right)$? Is it $$ P \left(X<\frac{2}{3},Y=\frac{1}{3} \right)=\int\int_D f(x,y)\,dx \,dy $$ where $D= \left\{ \left\{X<\frac{2}{3},Y=\frac{1}{3} \right\} \right \}\cap \{0\le x\le y\le 1\}\}$
You are on the right track. To find $P(X<\frac{2}{3}|Y=\frac{1}{3})$, you first need to compute the joint probability $P(X<\frac{2}{3},Y=\frac{1}{3})$ using the joint probability density function: \begin{align*} P(X<\frac{2}{3},Y=\frac{1}{3}) &= \iint_D f(x,y) , dxdy \\ &= \int_{1/3}^{2/3} \int_{0}^{2/3} 3y , dx , dy \\ &= \frac{1}{3}, \end{align*}
where $D$ is the region of integration given by $D = {(x,y) : 0 \leq x \leq \frac{2}{3}, \frac{1}{3} \leq y \leq 1}$.
Then, using the formula $P(X<\frac{2}{3}|Y=\frac{1}{3}) = \frac{P(X<\frac{2}{3},Y=\frac{1}{3})}{P(Y=\frac{1}{3})}$, we get: $$ P(X<\frac{2}{3},Y=\frac{1}{3}) = \int_{1/3}^{2/3} \int_{x}^{1/3} 3y \, dy \, dx = \frac{1}{18}. $$