2011 Cayley Problem 25

Two numbers a and b with 0 \le a \le 1 and 0 \le b \le 1 are chosen at random. The number c is defined by c = 2a + 2b. The numbers a, b and c are each rounded to the nearest integer to give A, B and C, respectively. (For example, if a = 0.432 and b = 0.5, then c = 1.864, and so A = 0, B = 1, and C = 2.) What is the probability that 2A + 2B = C?

\textbf{(A)}\ \frac{15}{32}\quad \textbf{(B)}\ \frac{3}{8}\quad \textbf{(C)}\ \frac{1}{2}\quad \textbf{(D)}\ \frac{7}{16}\quad \textbf{(E)}\ \frac{3}{4}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Cayley contests here.