2003 AMC 10B 第 21 题

Question

一个袋子里有两颗红珠和两颗绿珠。每次从袋中取出一颗珠子，不论取出的是什么颜色，都放回一颗红珠。这样替换三次后，袋中所有珠子都是红色的概率是多少？

A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?

\dfrac{1}{8}

\dfrac{5}{32}

\dfrac{9}{32}

\dfrac{3}{8}

\dfrac{7}{16}

答案：C

知识点：基本概率分类讨论

难度评级：1600

小提示：

最终全为红色，当且仅当三次抽取中两颗绿珠都被抽到。

All beads are red exactly when both green beads are drawn during the three draws

大提示：

分别计算抽取序列 GG、GRG、RGG 的概率，并跟踪袋中珠子数变化。

Add the probabilities of the draw sequences GG, GRG, and RGG, tracking how the counts change

解答：

袋中始终有

4

颗珠子。

最终全为红色，正好表示两颗绿珠都被抽到。先绿后绿的概率为

\dfrac24 \cdot \dfrac14 = \dfrac18

。绿、红、绿的概率为

\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}

。红、绿、绿的概率为

\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}

。

把三种情况相加，得到

\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}\text{。}

所以正确答案是 C。

The bag always holds

4

beads. All are red at the end precisely when both greens are drawn.

Drawing green then green has probability

\dfrac24 \cdot \dfrac14 = \dfrac18.

Green, red, green has probability

\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}.

Red, green, green has probability

\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}.

The total is

\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}.

Thus, the correct answer is C.

2003 AMC 10B 第 21 题

答案：C

难度评级：1600

小提示：

大提示：

解答：

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