AMC10 2010 A

AMC10 2010 A · Q23

AMC10 2010 A · Q23. It mainly tests Linear equations, Probability (basic).

Each of 2010 boxes in a line contains a single red marble, and for 1 ≤ k ≤ 2010, the box in the kth 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) < 1/2010?

有 2010 个盒子排成一行，每个盒子含一个红弹珠，对于 $1 \leq k \leq 2010$，第 $k$ 个盒子还含 $k$ 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。设 $P(n)$ 为 Isabella 恰好抽取 $n$ 颗弹珠后停止的概率。求 $P(n) < 1/2010$ 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

Answer (A): If Isabella reaches the $k^{\text{th}}$ box, she will draw a white marble from it with probability $\frac{k}{k+1}$. For $n\ge 2$, the probability that she will draw white marbles from each of the first $n-1$ boxes is $$ \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{n-1}{n}=\frac{1}{n}, $$ so the probability that she will draw her first red marble from the $n^{\text{th}}$ box is $P(n)=\frac{1}{n(n+1)}$. The condition $P(n)<1/2010$ is equivalent to $n^2+n-2010>0$, from which $n>\frac{1}{2}(-1+\sqrt{8041})$ and $(2n+1)^2>8041$. The smallest positive odd integer whose square exceeds 8041 is 91, and the corresponding value of $n$ is 45.

答案（A）：如果伊莎贝拉到达第 $k$ 个盒子，她从中抽到白色弹珠的概率为 $\frac{k}{k+1}$。当 $n\ge 2$ 时，她在前 $n-1$ 个盒子中都抽到白色弹珠的概率为 $$ \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{n-1}{n}=\frac{1}{n}, $$ 因此，她第一次抽到红色弹珠来自第 $n$ 个盒子的概率为 $P(n)=\frac{1}{n(n+1)}$。条件 $P(n)<1/2010$ 等价于 $n^2+n-2010>0$，由此得到 $n>\frac{1}{2}(-1+\sqrt{8041})$ 且 $(2n+1)^2>8041$。平方超过 8041 的最小正奇数是 91，对应的 $n$ 值为 45。

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