2021 AMC Fall 10A Problem 13

1. What "more than half" means. Each ball has $5$ other balls. More than half of $5$ means at least $3$. So every ball must have at least $3$ balls of the other color.
2. Try the possible color splits. Only the number of black balls matters. Go through the splits:

Only the split $3$ black, $3$ white works. With it, every ball differs from exactly $3$ of the other $5$.
3. Count all colorings. Each ball has $2$ choices, so there are $2^6 = 64$ equally likely colorings.
4. Count the good ones. Pick which $3$ of the $6$ balls are black: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways.
5. Probability. $\frac{20}{64} = \frac{5}{16}$.

The answer is $\boxed{\textbf{(D)}\ \frac{5}{16}}$.

A rule that must hold for every ball usually turns into a rule about the totals. The black balls push the number of whites up, and the white balls push the number of blacks up. With only $6$ balls, the two pushes meet at exactly $3$ and $3$. After that, it is one counting step over $64$.

For any one ball, the "other" balls number $5$, so "more than half" means at least $3$ of them have the opposite color.

Suppose $k$ balls are black and $6 - k$ are white. A black ball (if any) sees $6 - k$ white balls among the others, so we need $6 - k \ge 3$, i.e. $k \le 3$. A white ball (if any) sees $k$ black balls, so we need $k \ge 3$. Both colors must be present (if all six were black, each ball would differ from $0$ others), so both conditions apply and $k = 3$ exactly.

Conversely, with $3$ black and $3$ white, every ball differs from exactly $3$ of the other $5$, which is more than half. So the event is exactly "three of each color."

There are $2^6 = 64$ equally likely colorings, of which $\binom{6}{3} = 20$ have three black balls:
$$P = \frac{20}{64} = \frac{5}{16}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{5}{16}}$.