2016 AMC 10A Problem 17

The green balls all look alike, so an arrangement is decided by one thing: where the red ball sits.

1. There are $N + 1$ balls in the line, so the red ball can be in any of $N + 1$ spots, each equally likely. If $L$ green balls are to its left, then $L$ can be $0, 1, 2, \ldots, N$, each with chance $\frac{1}{N+1}$.
2. Try a small case, $N = 10$. We need at least $\tfrac35 \cdot 10 = 6$ greens on one side. That fails only when both sides have $5$ or fewer, which forces $L = 5$ exactly. So $1$ bad spot out of $11$, and $P(10) = \tfrac{10}{11}$.
3. In general write $N = 5m$, so we need $3m$ greens on one side. The rule fails when the left side has fewer than $3m$ and the right side, $5m - L$, has fewer than $3m$. The second condition means $L$ is more than $2m$. So the bad values are
$$L = 2m + 1,\ 2m + 2,\ \ldots,\ 3m - 1,$$
which is $m - 1$ values. (For $N = 10$, $m = 2$, this gives $1$ bad spot, as found above.)
4. So
$$P(N) = 1 - \frac{m - 1}{5m + 1}.$$
This matches the problem's checks: $m = 1$ gives $P = 1$, and for large $m$ the fraction is close to $\tfrac{m}{5m} = \tfrac15$, so $P$ is close to $\tfrac45$.
5. We want $P(N) < \tfrac{321}{400}$, which means the bad fraction is more than $1 - \tfrac{321}{400} = \tfrac{79}{400}$:
$$\frac{m - 1}{5m + 1} > \frac{79}{400}.$$
6. Cross-multiply (both denominators are positive): $400(m - 1) > 79(5m + 1)$, so $400m - 400 > 395m + 79$, which gives $5m > 479$, or $m > 95.8$.
7. The smallest whole number $m$ is $96$, so $N = 5 \cdot 96 = 480$. (Check $m = 95$: $\tfrac{94}{476}$ is just under $\tfrac{79}{400}$ because $94 \cdot 400 = 37600 < 79 \cdot 476 = 37604$.)
8. The digit sum of $480$ is $4 + 8 + 0 = 12$.

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

A long line of identical balls with one special ball has a tiny set of outcomes: just the special ball's spot. The bad spots form one unbroken block in the middle, so counting them is easy, but it is easy to be off by one. The problem hands you two checks ($P(5) = 1$ and the limit $\tfrac45$); test your formula against them before solving.

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 $\boxed{\textbf{(A)}\ 12}$.