AMC12 2016 A

AMC12 2016 A · Q13

AMC12 2016 A · Q13. It mainly tests Fractions, Probability (basic).

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 $P(N)<\frac{321}{400}$?

设 $N$ 为 $5$ 的正倍数。将 $1$ 个红球和 $N$ 个绿球按随机顺序排成一列。令 $P(N)$ 表示：至少有 $\frac{3}{5}$ 的绿球位于红球同一侧的概率。注意到 $P(5)=1$，且当 $N$ 趋于无穷大时，$P(N)$ 趋近于 $\frac{4}{5}$。求满足 $P(N)<\frac{321}{400}$ 的最小 $N$ 的各位数字之和。

(A) 12 12

(B) 14 14

(D) 18 18

(E) 20 20

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $N=5k$, where $k$ is a positive integer. There are $5k+1$ equally likely possible positions for the red ball in the line of balls. Number these $0,1,2,3,\ldots,5k-1,5k$ from one end. The red ball will not divide the green balls so that at least $\frac{3}{5}$ of them are on the same side if it is in position $2k+1,2k+2,\ldots,3k-1$. This includes $(3k-1)-2k=k-1$ positions. The probability that $\frac{3}{5}$ or more of the green balls will be on the same side is therefore $$1-\frac{k-1}{5k+1}=\frac{4k+2}{5k+1}.$$ Solving the inequality $\frac{4k+2}{5k+1}<\frac{321}{400}$ for $k$ yields $k>\frac{479}{5}=95\frac{4}{5}$. The value of $k$ corresponding to the required least value of $N$ is therefore $96$, so $N=480$. The sum of the digits of $N$ is $12$.

答案（A）：设 $N=5k$，其中 $k$ 为正整数。红球在一排球中的可能位置共有 $5k+1$ 个，且等可能。从一端起将这些位置编号为 $0,1,2,3,\ldots,5k-1,5k$。当红球位于位置 $2k+1,2k+2,\ldots,3k-1$ 时，它不会把绿球分成使得至少有 $\frac{3}{5}$ 的绿球位于同一侧的情形。该区间包含 $(3k-1)-2k=k-1$ 个位置。因此，$\frac{3}{5}$ 或更多绿球在同一侧的概率为 $$1-\frac{k-1}{5k+1}=\frac{4k+2}{5k+1}.$$ 解不等式 $\frac{4k+2}{5k+1}<\frac{321}{400}$ 得 $k>\frac{479}{5}=95\frac{4}{5}$。满足所需的最小 $N$ 对应的 $k$ 为 $96$，所以 $N=480$。$N$ 的各位数字和为 $12$。

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