Question

设 $N$ 是 $5$ 的正倍数。将一个红球和 $N$ 个绿球随机排成一行。令 $P(N)$ 为至少有 $\frac{3}{5}$ 的绿球在红球同一侧的概率。注意 $P(5)=1$ ，且当 $N$ 变大时 $P(N)$ 趋近于 $\frac{4}{5}$ 。使 $P(N) < \dfrac{321}{400}$ 的最小 $N$ 的各位数字之和是多少？

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 $\frac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\frac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \dfrac{321}{400}?$

$12$

$14$

$16$

$18$

$20$

答案：A

知识点：基本概率不等式

难度评级：1790

解答：

先排好 $N$ 个绿球，再把红球放入 $N+1$ 个空隙之一。若红球左侧有 $k$ 个绿球，则右侧有 $N-k$ 个。

至少有 $\frac35N$ 个绿球在一侧，等价于 $k\le\frac25N$ 或 $k\ge\frac35N$ 。因此不满足条件的空隙为共 $\frac15N-1$ 个。 $\frac25N+1,\frac25N+2,\ldots,\frac35N-1,$

所以解不等式 $\frac{4N+10}{5N+5}<\frac{321}{400}$ 得 $N>479$ 。最小的 $5$ 的正倍数是 $480$ ，其各位数字之和为 $12$ 。 $\begin{aligned} P(N) &= 1-\frac{N/5-1}{N+1} \\ &= \frac{4N+10}{5N+5}. \end{aligned}$

所以正确答案是 A。

Think of first arranging the $N$ green balls, then placing the red ball in one of the $N+1$ gaps. If $k$ green balls are to the left of the red ball, then $N-k$ are to its right.

At least $\frac35N$ green balls are on one side exactly when $k\le\frac25N$ or $k\ge\frac35N$ . Thus the bad gaps are $\frac25N+1,\frac25N+2,\ldots,\frac35N-1,$ a total of $\frac15N-1$ gaps.

Therefore $\begin{aligned} P(N) &= 1-\frac{N/5-1}{N+1} \\ &= \frac{4N+10}{5N+5}. \end{aligned}$ Solving $\frac{4N+10}{5N+5}<\frac{321}{400}$ gives $N>479$ . The least positive multiple of $5$ is $480$ , whose digit sum is $12$ .

Thus, the correct answer is A.

2016 AMC 10A 第 17 题

答案：A

难度评级：1790

解答：

其他年份的第 17 题