Joint Probability density function given as follows $f(x,y)=30xy^2$ if $ x-1\leq y\leq 1-x , 0\leq x\leq 1$ [closed]

Joint Probability density function given as follows $f(x,y)=30xy^2$ if $x-1\leq y\leq 1-x$ , $\,\,0\leq x\leq 1$

Find $P(X>Y)$

I sketched the region of integration and i have found that the vertices are $(1/2,1/2),(0,0),(0,-1)$ and $(1,0)$ however i still cant find the correct answer after I did the double integration. The answer given is $\frac{21}{32}$

Solution 1:

Since you have sketched the graph....you will need to integrate over this region to get $P(X<Y)$. Then you subsract it from $1$ to get the desired probability.

Here is the sketch(don't laugh at it...I'm poor at drawing).

As you can see from the picture, the integral is

$$\int_{0}^{\frac{1}{2}}\int_{x}^{1-x}30xy^{2}\,dy\,dx=\frac{11}{32}$$.

So $P(X>Y)=1-\frac{11}{32}=\frac{21}{32}$