AMC10 2003 B

AMC10 2003 B · Q21

AMC10 2003 B · Q21. It mainly tests Probability (basic), Casework.

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?

一个袋子里有2颗红色珠子和2颗绿色珠子。你伸手进袋子取出颗珠子，无论取出的是什么颜色的珠子，都用一颗红色珠子替换回去。经过三次这样的替换后，袋子里所有珠子都是红色的概率是多少？

(A) 1/8 1/8

(B) 5/32 5/32

(D) 3/8 3/8

(E) 7/16 7/16

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) The beads will all be red at the end of the third draw precisely when two green beads are chosen in the three draws. If the first bead drawn is green, then there will be one green and three red beads in the bag before the second draw. So the probability that green beads are drawn in the first two draws is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$. The probability that a green bead is chosen, then a red bead, and then a green bead, is $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{32}$. Finally, the probability that a red bead is chosen then two green beads is $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{16}$. The sum of these probabilities is $\frac{1}{8}+\frac{3}{32}+\frac{1}{16}=\frac{9}{32}$.

（C）在第三次抽取结束时，珠子全为红色当且仅当三次抽取中恰好抽到了两颗绿色珠子。若第一次抽到的是绿色珠子，则在第二次抽取前袋中将有一颗绿色珠子和三颗红色珠子。因此，前两次抽取都抽到绿色珠子的概率为 $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$。先抽到一颗绿色珠子，再抽到一颗红色珠子，然后抽到一颗绿色珠子的概率为 $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{32}$。最后，先抽到一颗红色珠子，然后再抽到两颗绿色珠子的概率为 $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{16}$。这些概率之和为 $\frac{1}{8}+\frac{3}{32}+\frac{1}{16}=\frac{9}{32}$。

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