Show that $P(X
Let $X$ and $Y$ be nonnegative, independent continuous random
  variables. 
(a) Show that $$P(X<Y) = \int _0^\infty F_X(x)f_Y(x) \ dx$$ 
  (b) What does this become if $X ∼ \exp(λ_1)$ and $Y ∼ \exp(λ_2)$?
I don't understand what exactly $P(X<Y)$ means, except for "the probability that the random variable $X$ is less than the random variable $Y$". I think I might be able to solve (b) myself but for (a) I don't know where to start.
 A: Your interpretation of $\mathbb P(X < Y)$ is correct -- the next step is turning that idea into a mathematical expression you can work with.
If you have two random variables with joint density $f(x,y)$, then $\mathbb P(X < Y)$ can be computed as
$$\mathbb P(X < Y) = \iint_{\{x < y\}} f(x, y) \, \textrm d x \, \textrm d y.$$
Since these variables are supported on $[0, \infty)$, one could also write this as
$$\int_0^{\infty} \int_0^y f(x, y) \, \textrm dx \, \textrm dy.$$
But in the case where $X, Y$ are independent, the joint density $f(x,y)$ is just a product of their individual densities. Can you take it from here? (I can provide more hints if not.)
A: In modern probability, conditional probability is define on special case of conditional expectation.
so define:
$A=\{\omega \in \Omega| X(\omega) < Y(\omega)\}$
$p(X< Y)= p(A)=E(I_A)=E(E(I_A|Y))=\int E(I_A|Y=y) f_Y(y) dy=\int E(I_{X< Y}|Y=y) f_Y(y) dy=\int E(I_{X< y}|Y=y) f_Y(y) dy\overset{independent}{=}
\int E(I_{X< y}) f_Y(y)dy=\int P(X< y) f_Y(y)dy
=\int F_X(y) f_Y(y)dy$
Note that $E(I_A|X)=E(Z|X)=E(Z|\sigma(X))$ is a predictor that satisfy
projection property and exists and unique. 
