A random point $(X,Y)$ is distributed uniformly on the triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$. For this random point, what is $P(X > a)$ where $0 < a < 1$?
Attempted solution:
The joint pdf $f(x,y) = c$ thus when I integrate $$\int_{0}^{1}\int_{0}^{1-x}c dy dx = 1$$ we get $c = 2$. Thus the boundaries are $0 < x < 1$ and $0 < y < 1-x$. I am just not sure how to calculate the $P(X > a)$.
Any suggestions are greatly appreciated.
The joint PDF of $(X,Y)$ can be written as: $$f_{X,Y}(x,y)=\cases{2,\ 0\leq x\leq 1,0\leq y\leq 1-x\\0,\ \text{anywhere else}}$$We may introduce the marginal distribution of $X$, which is $$f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)dy=\int_{0}^{1-x} 2dy=2(1-x)$$ for $0<x<1$ and $0$ for any $x$ outside this range. To find the probability that $X>a$ we simply integrate $f_X(x)$ from $a$ to 1 $$\Pr(X>a)=\int_a^1f_X(x)dx=\int_a^12(1-x)dx\\=2(1-a)-(1-a^2)=a^2-2a+1$$