Select numbers $a$ and $b$ between $0$ and $1$ independently and at random, and let $c$ be their sum. Let $A, B$ and $C$ be the results when $a, b$ and $c$, respectively, are rounded to the nearest integer. What is the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that $A + B = C$?
$\text {(A)}\ \frac14 \qquad \text {(B)}\ \frac13 \qquad \text {(C)}\ \frac12 \qquad \text {(D)}\ \frac23 \qquad \text {(E)}\ \frac34$
Solution 1
[Casework](https://artofproblemsolving.com/wiki/index.php/Casework):

$0 + 0 = 0$. The probability that $a  \frac 12$ is $\frac 14$, but now $\frac{1}{2}  3/2$ then this will work because in order for this to happen, $a$ and $b$ are each greater than $1/2$ making $A = B = 1$, and $C = 2$. Each of these triangles in the unit square has area of 1/8.
The only case left is when $C = 1$. Then each of $A$ and $B$ must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.
Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is $\frac 34$.