Probability that A < B I would like to know if  I had a line of length 1 (think of the real number line from 0 to 1), and I were to randomly place 2 points on the line, what is the probability that point B is right of point A.
I think the probability is 1/2. 
If A was on 0, the probability of B being placed correctly is 100%. 
If A was on 1/4, the probability of B being placed correctly is 75%.
If A was on 1/2, the probability of B being placed correctly is 50%.
If A was on 3/4, the probability of B being placed correctly is 25%.
If A was on 1, the probability of B being placed correctly is 0%.
This led me to consider this integral
$$\int_0^1 (-x+1)dx$$
where 1 * dx represents the probability of A being on a certain point on the number line and (-x + 1) representing the probability of B being right of A at that particular point. The integral sums up all of these products.
Is this correct??
 A: The symmetry argument gets you there faster, but there is nothing wrong with the approach you are taking. (And it generalizes to the case where $A$ and $B$ are not identically distributed.) Your argument is to condition on the value of $A$: if $A=x$, then the probability that $B$ exceeds $x$ is $1-x$, as you've argued. To get the final answer you average out this conditional probability over the density of $A$:
$$P(B>A)=\int P(B>A\mid A=x)f_A(x)\,dx=\int_0^1 (1-x)\cdot 1\,dx. $$ In the final step we plug in the density of $A$, which equals $1$ for $x$ between $0$ and $1$.
A: I think I see the point that you want to verify. The way you model this problem is as follows, 
$$P(\text{B is right of A} ) = \sum_{\text{for all x in the line}}P(\text{B is right of x})P(\text{A is on x})$$
Now in order to show clearly how your method works, I assume the length of the Interval to be $L$ (instead of 1). Hence the probability of $A$ on $x$ is $\frac{1}{L}$, the probability of $B$ being right of $A$ is $\frac{L-x}{L}$
$$P(\text{B is right of A}) = \int_{0}^{L}P(\text{B is right of x})P(\text{A is on x}) dx$$
$$P(\text{B is right of A}) = \int_{0}^{L}\frac{L-x}{L}\frac{1}{L} dx$$
$$P(\text{B is right of A}) = \frac{1}{2}$$
A: If you want a formal answer to this question, consider the measurable space $\Omega=[0,1]^2$ equipped with its Borel-$\sigma$-algebra and the probability measure $\mathbb{P}=\lambda \otimes \lambda$ (the product Lebesgue-measure on $\Omega$). Clearly, this probability measure corresponds to the uniform distribution on $\Omega$.
Furthermore denote $X=(X_1,X_2): \Omega \to \Omega$ the identity map. Then, 
$$ \mathbb{P}(X_1<X_2) \; = \; \int_{[0,1]} \int_{[0,1]} \mathbb{1}_{\{x_1<x_2\}}\; dx_1 dx_2 \; = \; \int_{[0,1]} \int_{0}^{x_2} 1\; dx_1 dx_2 \; = \; \frac12$$
