Q: Comparing probabilty density functions A is generated randomly from probability density function e-e^(1-x)
B is generated randomly from probability density function 3x^2
by looking at the graphs you can see that B is going to be greater than A the majority of the time
my question is what's the probability that B will be greater than A?
80%? 90%?
(btw I hope I have not misunderstood what a probability density function is)

A: In order to solve the problem, we will need to assume that the random variables $A$ and $B$ are independent.
It is convenient to will call $A$ by the name $X$, and $B$ by the name $Y$. The density function of $X$. The density function of $X$ is $e-e^{1-x}$ on the interval $0\le x\le 1$, and $0$ elsewhere.  
The density function of $Y$ is 
is $3y^2$ in the interval $0\le y\le 1$, and $0$ elsewhere.  
By independence, the joint density of $X$ and $Y$ is 
$$\left(e-e^{1-x}\right)(3y^2)$$
on the unit square, and $0$ elsewhere. 
The probability that $Y\gt X$ is the integral over the part of the unit square that has $y\gt x$ of the joint density function.
One can evaluate the integral in various ways. It is doable but somewhat messy.
 For example, we can integrate first with respect to $x$, then with respect to $y$. We get 
$$\int_{y=0}^1 \int_{x=0}^y \left(e-e^{1-x}\right)(3y^2)\,dx\,dy.$$
