Probability: One red ball and $N$ green balls (AMC 10 question) Let $N$ be a positive multiple of $5$. One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\frac35$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\frac45$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \frac{321}{400}$?
This is question from AMC 10, however, I am lost with their explanations. Thanks.

Since $N$ is a multiple of $5$, let $N = 5n$ for some positive integer $n$.  Then we have a total of $5n + 1$ balls.  Consider the complementary event that less than $3/5$ of the green balls are on each side of the red ball, which suggests that the red ball is "near the center" of the line.
In particular, we can see that if there exists a side for which at least $3n$ (green) balls are on that side of the red ball, then the red ball is "too far away" from the center.  Specifically, if we number the balls $1, 2, \ldots, 5n+1$, the largest number we can pick for the red ball is $3n$, since if we pick $3n+1$ or greater, then the balls $1, 2, \ldots, 3n$ are all green.  Similarly, the smallest number we can pick for the red ball is $(5n+1) - (3n-1) = 2n+2$, since if we pick $2n+1$ or less, the balls $2n+2, \ldots, 5n+1$ are all green.  It follows that we can only pick the numbers $$\{2n+2, \ldots, 3n\}.$$  If $n = 1$, this is the empty set, since $4 > 3$.  For $n > 1$, there are $$3n - (2n+2) + 1 = n-1$$ permissible choices of the red ball out of $5n+1$ choices, therefore the complementary probability is $$1 - \Pr[N] = \frac{n-1}{5n+1},$$ and the desired probability is $$\Pr[N] = \frac{4n+2}{5n+1}.$$  Solving the inequality $$\frac{4n+2}{5n+1} < \frac{321}{400},$$ we get $400(4n+2) < 321(5n+1)$ or $5n > 479$, or $n > 95.8$.  The smallest such integer satisfying this inequality is $96$, hence $N = 5(96) = 480$. 
