I wish to double check my answers so that I know how to handle bounds when the region of $x$ and $y$ is a triangle.
a. Are $X$ and $Y$ independent?
No because the region of x and y. $0 \leq x, 0 \leq y, x+y < 1$ forms a triangle so we can automatically tell that it's dependent as the region must be rectangular to even be considered independent.
b. Find $P(X < 1/2)$.
I found that the density is $f(x) = \int_0^{1-x}1512x^2y^5dy = 252x^2(1-x)^6$ for $0 \leq x <1$ so $P(X < 1/2) = \int_0^{1/2}252x^2[1-x]^6 dx = 233/256$
c. Find $P(Y < 1/2)$.
I found that the density is $f(y) = \int_0^{1-y}1512x^2y^5dx = 504y^5(1-y)^3$ for $0\leq y < 1$ So $P(Y< 1/2) = \int_0^{1/2} 504y^5(1-y)^3 dy = 65/256 $
d. Find $P(X + Y < 1/2)$.
$P(X+Y < 1/2) = \int_0^{1/2}\int_0^{\cfrac{1}{2}-y}1512x^2y^5dxdy = 1/512$
I just wish to double check if my methodology and bounds of integration are correct. When I draw a picture I think I got the correct answers but I just want to make sure.
Thank you for any help!
Your integrals look correct. Just a few comments.
a. Alternatively, you can find the marginal density functions $f_X$ and $f_Y$ and compare the product $f_X(x)f_Y(y)$ with $f(x,y)$.
bc. You should not use the same symbol $f$ for both the density functions $X$ and $Y$ (and the joint density). To avoid writing the number $1512$ over and over again, you can introduce $A=1512$, and write the joint density as $f(x,y)=Ax^2y^5$. You can then substitute the number in the last step.
The density of $X$ is $$ f_X(x)=\int_{0}^{1-x}f(x,y)dy=A\int_{0}^{1-x}x^2y^5,\quad 0\le x\le 1 $$ and the density of $Y$ is $$ f_Y(y)=\int_0^{1-y}f(x,y)dx=A\int_0^{1-y}x^2y^5,\quad 0\le y\le 1 $$
Once you find $f_X$ and $f_Y$, you then have $$ P(X<1/2)=\int_{-\infty}^\infty f_X(x)dx=\int_0^{1/2}f_X(x)dx $$ $$ P(Y<1/2)=\int_0^{1/2}f_Y(y)dy $$
d. For the last question $$ P(X+Y<1/2)=\int_{0}^{1/2}\left(\int_0^{1/2-y}Ax^2y^5dx\right)dy $$