Originally Posted by

**Canadian0469** Hi all,

I'm trying to do the following problem and I'm having trouble figuring out how to set up my integrals:

Suppose that $\displaystyle Y_1$ and $\displaystyle Y_2$ are uniformly distributed over the triangle defined by the following points on y1,y2:

(-1,0), (0,1), (1,0)

a) Find $\displaystyle P(Y_1\leq{3/4}, Y_2\leq{3/4})$

What I've been able to do up to the point I'm confused:

Because it's uniformly distributed over a triangle, I found that the probability is uniformly distributed over area: bh/2 = 2*1/2 = 1

Therefore, I get the distribution function:

$\displaystyle f(y_1, y_2) = \begin{cases}1&\text{ if -1}\leq{y_1}\leq{1}, 0\leq{y_2}\leq1 \\0 &\text{otherwise}\end{cases}$

Now I can't figure out how to write the double-integral, can I just use one or do I have to find the sum of different components of the part of the triangle that fits the constraints?