Question

一个罐子里有红、白、蓝、绿四种颜色的弹珠。不放回地抽取四颗弹珠时，下列事件的发生概率相同：

（a）抽到四颗红色弹珠；

（b）抽到一颗白色弹珠和三颗红色弹珠；

（c）抽到一颗白色、一颗蓝色和两颗红色弹珠；以及

（d）每种颜色各抽到一颗弹珠。

满足上述条件的弹珠总数最少是多少？

An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(a) the selection of four red marbles;

(b) the selection of one white and three red marbles;

(c) the selection of one white, one blue, and two red marbles; and

(d) the selection of one marble of each color.

What is the smallest number of marbles satisfying the given condition?

$19$

$21$

$46$

$69$

大于 $69$

more than $69$

答案：B

知识点：组合 Diophantine equations

难度评级：2190

小提示：

设四种颜色的弹珠数分别为 $r,w,b,g$ ，令四个事件的组合数相等

Let the four color counts be $r,w,b,g$ and equate the combination counts for the four events

大提示：

逐次取比值，把 $w,b,g$ 表示成 $r$ 的函数，再求使三者都是整数的最小 $r$

Successive ratios determine $w,b,g$ in terms of $r$ ; then find the smallest $r$ making all three integers

解答：

相等的概率具有相同的总样本数，因此有利选法数满足 $\binom r4=w\binom r3$ $=wb\binom r2=wbgr$ 。逐次取比值得 $w=\frac{r-3}{4}$ 、 $b=\frac{r-2}{3}$ 和 $g=\frac{r-1}{2}$ 。使三者均为正整数的最小 $r\ge4$ 是 $r=11$ 。此时 $(w,b,g)=(2,3,5)$ ，弹珠总数为 $11+2+3+5=21$ 。所以正确答案是 B。

Equal probabilities have the same common denominator, so their favorable selection counts satisfy $\binom r4=w\binom r3$ $=wb\binom r2=wbgr.$ Successive ratios give $w=\frac{r-3}{4},$ $b=\frac{r-2}{3},$ and $g=\frac{r-1}{2}.$ The least $r\ge4$ making all three positive integers is $r=11.$ Then $(w,b,g)=(2,3,5),$ for $11+2+3+5=21$ marbles. Thus the correct answer is B.

1996 AMC 12 第 26 题

答案：B

难度评级：2190

小提示：

大提示：

解答：

其他年份的第 26 题