Question

一个盒子里有 $5$ 个筹码，编号为 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $5$ 。每次随机抽出一个筹码且不放回，直到抽出的编号之和超过 $4$ 。需要抽 $3$ 次的概率是多少？

A box contains $5$ chips, numbered $1,$ $2,$ $3,$ $4,$ and $5.$ Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4.$ What is the probability that $3$ draws are required?

$\dfrac{1}{15}$

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

答案：D

知识点：无放回抽样基本概率

难度评级：1290

解答：

恰好需要第三次抽取，当且仅当前两个筹码的和仍不超过 $4$ 。这样的无序对只有 $\{1,2\}$ 和 $\{1,3\}$ 。每一对有种抽取顺序，所以有 $4$ 个有利序列。

前两次有序抽取共有 $5\cdot4=20$ 种，所以概率为 $4/20=1/5$ 。正确答案是 D。

We need a third draw exactly when the first two chips still sum to $4$ or less. The only such unordered pairs are $\{1,2\}$ and $\{1,3\}.$ Each can be drawn in either order, giving $4$ favorable ordered prefixes.

Imagine that a complete random ordering of all five chips is chosen in advance. Then all $5\cdot4=20$ ordered first-two-chip prefixes are equally likely, even when the actual process would stop after the first chip. Thus the probability is $4/20=1/5.$ Therefore, the answer is D.

2018 AMC 10B 第 6 题

答案：D

难度评级：1290

解答：

其他年份的第 6 题