Question

A 队和 B 队进行系列赛，先赢三场的队伍赢得系列赛。每场比赛两队获胜概率相同，没有平局，且各场结果相互独立。已知 B 队赢了第二场，且 A 队赢得系列赛，那么 B 队赢第一场的概率是多少？

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

答案：A

知识点：条件概率系统列举

难度评级：1860

小提示：

想象五场比赛都进行，即使系列赛已经提前结束，这样每个五场序列等可能。

Imagine all five games are played, even after the series is decided, so each five-game sequence is equally likely

大提示：

列出 B 赢第 $2$ 场且 A 赢得系列赛的序列，再看其中多少个第 $1$ 场也是 B 赢。

List the sequences in which B wins game $2$ and A wins the series, then see how many also have B winning game $1$

解答：

假设五场都进行，则所有五场结果序列等可能。要求 B 赢第 $2$ 场且 A 最终以三场胜利赢得系列赛，得到以下等可能序列：

$\begin{gathered} \text{BBAAA}, \quad \text{ABBAA}, \\ \text{ABABA}, \quad \text{ABAAB}, \\ \text{ABAAA} \end{gathered}\text{。}$

其中只有 BBAAA 的第一场是 B 赢，所以概率为 $\dfrac{1}{5}$ 。

所以正确答案是 A。

Suppose all five games are played, so every sequence of five results is equally likely. Requiring that B wins game $2$ and A ends up with the series (three wins) leaves the equally likely sequences

$\begin{gathered} \text{BBAAA}, \quad \text{ABBAA}, \\ \text{ABABA}, \quad \text{ABAAB}, \\ \text{ABAAA}. \end{gathered}$

Only in BBAAA does team B win the first game, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is A.

2005 AMC 10A 第 18 题

答案：A

难度评级：1860

小提示：

大提示：

解答：

其他年份的第 18 题