Problem:
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?
$\textbf{(A) }\dfrac1{64}\qquad\textbf{(B) }\dfrac16\qquad\textbf{(C) }\dfrac14\qquad\textbf{(D) }\dfrac5{16}\qquad\textbf{(E) }\dfrac12$
Solution:
We perform casework on the number of white balls (there could be 1, 2, 3, 4, 5, or 6). However, notice that the only possible case which satisfies the given condition is when there are $3$ white balls and $3$ black balls, and there are $\binom{6}{3}=20$ ways for this to happen. There are $2^6$ ways in total, so the final probability is $$\frac{20}{64}=\frac{5}{16} \longrightarrow \boxed{\textbf{(D)}}$$
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