2016 AMC 10A Problem 17

The arrangement is determined by where the red ball sits among the $N+1$ positions, and each position is equally likely. If $L$ green balls are to its left, then $N - L$ are to its right, and $L$ takes each value $0, 1, \dots, N$ with probability $\frac{1}{N+1}$.

Write $N = 5m$. The condition fails when neither side holds at least $3m$ greens, i.e. $L < 3m$ and $5m - L < 3m$, which means $2m < L < 3m$. The failing values are $L = 2m+1, \dots, 3m-1$, a total of $m - 1$ positions. Hence
$$P(N) = 1 - \frac{m-1}{5m+1}.$$
(Check: $m = 1$ gives $P = 1$, and as $m \to \infty$ the fraction tends to $\frac15$, so $P \to \frac45$.)

We need $\frac{m-1}{5m+1} > 1 - \frac{321}{400} = \frac{79}{400}$:
$$400(m-1) > 79(5m+1) \iff 400m - 400 > 395m + 79 \iff 5m > 479 \iff m \ge 96.$$

So the least $N$ is $5 \cdot 96 = 480$, whose digit sum is $4 + 8 + 0 = 12$.

The answer is $\boxed{\textbf{(A)}\ 12}$.

A random line of identical green balls plus one red ball has a tiny sample space in disguise: only the red ball's slot matters, and all $N+1$ slots are equally likely. Counting the bad slots is easier than the good ones because they form one contiguous block. The problem's own sanity checks ($P(5) = 1$, limit $\frac45$) are there to catch off-by-one errors in that count; use them.

• Off-by-one errors in the count of failing positions (using m or m+1 instead of m-1) or forgetting that there are N + 1 positions, not N.
• Reporting $N = 480$ or $m = 96$ instead of the digit sum $12$.