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

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

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

解答：

其他年份的第 21 题