AMC12 2010 A

AMC12 2010 A · Q19

AMC12 2010 A · Q19. It mainly tests Quadratic equations, Probability (basic).

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$, the box in the $k\text{th}$ 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 $P(n) < \frac{1}{2010}$?

一排 2010 个盒子，每个盒子含有一个红弹珠，并且对 $1 \le k \le 2010$，第 $k\text{th}$ 个盒子还包含 $k$ 个白弹珠。Isabella 从第一个盒子开始，依次从每个盒子随机抽取一颗弹珠。她在第一次抽到红弹珠时停止。令 $P(n)$ 为 Isabella 恰好抽取 $n$ 颗弹珠后停止的概率。求使得 $P(n) < \frac{1}{2010}$ 的最小 $n$ 值。

(A) 45 45

(B) 63 63

(D) 201 201

(E) 1005 1005

Answer

Correct choice: (A)

正确答案：（A）

Solution

The probability of drawing a white marble from box $k$ is $\frac{k}{k + 1}$, and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$. To stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \ldots, n-1,$ and draw a red marble from box $n.$ Thus, \[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\] So, we must have $\frac{1}{n(n + 1)} < \frac{1}{2010}$ or $n(n+1) > 2010.$ Since $n(n+1)$ increases as $n$ increases, we can simply test values of $n$; after some trial and error, we get that the minimum value of $n$ is $\boxed{\textbf{(A) }45}$, since $45(46) = 2070$ but $44(45) = 1980.$

从第 $k$ 个盒子抽到白弹珠的概率是 $\frac{k}{k + 1}$，从第 $k$ 个盒子抽到红弹珠的概率是 $\frac{1}{k+1}$。要在抽取了 $n$ 颗弹珠后停止，必须在盒子 $1, 2, \ldots, n-1$ 中都抽到白弹珠，并在盒子 $n$ 中抽到红弹珠。因此 \[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\] 所以需要 $\frac{1}{n(n + 1)} < \frac{1}{2010}$，即 $n(n+1) > 2010$。由于 $n(n+1)$ 随 $n$ 增大而增大，直接试值可得最小的 $n$ 为 $\boxed{\textbf{(A) }45}$，因为 $45(46) = 2070$ 而 $44(45) = 1980$。

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