Each of $6$ balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other $5$ balls?
$\textbf{(A) } \frac{1}{64}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{5}{16}\qquad\textbf{(E) }\frac{1}{2}$
Note that for this restriction to be true, there must be $3$ balls of each color. There are a total of $2^6 = 64$ ways to color the balls, and there are ${6 \choose 3} = 20$ ways for three balls chosen to be painted white.  Thus, the answer is $\frac{20}{64} = \boxed{\textbf{(D) } \frac{5}{16}}$. e
-Aidensharp