Question

一排有 $2010$ 个盒子，每个盒子里有一颗红弹珠；对 $1 \le k \le 2010$ ，第 $k$ 个盒子还含有 $k$ 颗白弹珠。Isabella 从第一个盒子开始，依次从每个盒子中随机抽取一颗弹珠。她第一次抽到红弹珠时停止。令 $P(n)$ 为 Isabella 正好抽取 $n$ 颗弹珠后停止的概率。使 $P(n) \lt \dfrac{1}{2010}$ 的最小 $n$ 是多少？

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010,$ the box in the $k$ th position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) \lt \dfrac{1}{2010}?$

$45$

$63$

$64$

$201$

$1005$

答案：A

知识点：基本概率裂项相消不等式

难度评级：2240

解答：

第 $k$ 个盒子共有 $k + 1$ 颗弹珠，所以从中抽到白弹珠的概率为 $\dfrac{k}{k + 1}$ 。

抽到红弹珠的概率为 $\dfrac{1}{k + 1}$ 。要在第 $n$ 次抽取后停止，前 $n - 1$ 次必须抽到白弹珠，最后一次抽到红弹珠。

概率为 $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$

分子与相邻分母相消，得到概率为 $\dfrac{1}{n(n + 1)}$ 。

需要找出满足的最小 $n$ 。 $\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$ $n(n + 1) \gt 2010.$

最小满足条件的 $n$ 为 $45$ 。

所以正确答案是 A。

Since there are $k + 1$ marbles in the $k$ th box, there is a $\dfrac{k}{k + 1}$ chance Isabella draws a white marble from it.

The probability of drawing a red marble is then $\dfrac{1}{k + 1}.$ To stop after drawing the $n$ th marble, the first $n - 1$ marbles must have been white.

This happens with a probability of $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to $\dfrac{1}{n(n + 1)}.$

We have to find the smallest $n$ such that $\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$ $n(n + 1) \gt 2010.$

Guessing and checking gives us that the smallest $n$ that works is $45.$

Thus, A is the correct answer.

2010 AMC 10A 第 23 题

答案：A

难度评级：2240

解答：

其他年份的第 23 题