AMC10 2016 A

AMC10 2016 A · Q17

AMC10 2016 A · Q17. 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$N$个绿球以随机顺序排成一列。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$P(5)=1$，并且当$N$趋于很大时，$P(N)$趋近于$\frac{4}{5}$。求使得$P(N)<\frac{321}{400}$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。红球在球列中有 $5k+1$ 个等可能的位置。将这些位置从一端编号为 $0,1,2,3,\ldots,5k-1,5k$。若红球位于位置 $2k+1,2k+2,\ldots,3k-1$，则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，至少有 $\frac{3}{5}$ 的绿球位于同一侧的概率为 $1-\frac{k-1}{5k+1}=\frac{4k+2}{5k+1}$。解不等式 $\frac{4k+2}{5k+1}<\frac{321}{400}$ 得 $k>\frac{479}{5}=95\frac{4}{5}$。对应所需的最小 $N$ 的 $k$ 值为 $96$，所以 $N=480$。$N$ 的各位数字之和为 $12$。

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