AMC12 2004 A

Casework: 1. $0 + 0 = 0$. The probability that $a < \frac{1}{2}$ and $b < \frac{1}{2}$ is $\left(\frac 12\right)^2 = \frac{1}{4}$. Notice that the sum $a+b$ ranges from $0$ to $1$ with a symmetric distribution across $a+b=c=\frac 12$, and we want $c < \frac 12$. Thus the chance is $\frac{\frac{1}{4}}2 = \frac 18$. 2. $0 + 1 = 1$. The probability that $a < \frac 12$ and $b > \frac 12$ is $\frac 14$, but now $\frac{1}{2} < a+b = c < \frac 32$, which makes $C = 1$ automatically. Hence the chance is $\frac 14$. 3. $1 + 0 = 1$. This is the same as the previous case. 4. $1 + 1 = 2$. We recognize that this is equivalent to the first case. Our answer is $2\left(\frac 18 + \frac 14 \right) = \frac 34 \Rightarrow \mathrm{(E)}$. Use areas to deal with this continuous probability problem. Set up a unit square with values of $a$ on x-axis and $b$ on y-axis. If $a + b < 1/2$ then this will work because $A = B = C = 0$. Similarly if $a + b > 3/2$ 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 $\frac 34$.