AMC10 2005 A

AMC10 2005 A · Q18

AMC10 2005 A · Q18. It mainly tests Probability (basic), Conditional probability (basic).

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?

A队和B队进行一系列比赛，先赢三场的队伍赢得系列赛。每场比赛两队获胜概率相等，无平局，各场比赛结果独立。如果B队赢得第二场比赛且A队赢得系列赛，则B队赢得第一场比赛的概率是多少？

(A) $\frac{1}{5}$ $\frac{1}{5}$

(B) $\frac{1}{4}$ $\frac{1}{4}$

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) $\frac{2}{3}$ $\frac{2}{3}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) There are four possible outcomes, ABAA, ABABA, ABBAA, and BBAAA, but they are not equally likely. This is because, in general, the probability of any specific four-game series is $(1/2)^4 = 1/16$, whereas the probability of any specific five-game series is $(1/2)^5 = 1/32$. Thus the first listed outcome is twice as likely as each of the other three. Let $p$ be the probability of the occurrence $ABBAA$. Then the probability of $ABABA$ is also $p$, as is the probability of $BBAAA$, whereas the probability of $ABAA$ is $2p$. So $$2p + p + p + p = 1,\quad \text{and}\quad p = \frac{1}{5}.$$ The only outcome in which team B wins the first game is $BBAAA$, so the probability of this occurring is $1/5$.

（A）有四种可能的结果： ABAA、ABABA、ABBAA 和 BBAAA，但它们并非等可能。这是因为，一般而言，任何一个特定的四场比赛系列发生的概率是 $(1/2)^4 = 1/16$，而任何一个特定的五场比赛系列发生的概率是 $(1/2)^5 = 1/32$。因此，列出的第一个结果发生的可能性是其他三个中每一个的两倍。令 $p$ 表示出现 $ABBAA$ 的概率。则出现 $ABABA$ 的概率也为 $p$，出现 $BBAAA$ 的概率也为 $p$，而出现 $ABAA$ 的概率为 $2p$。因此 $$2p + p + p + p = 1,\quad \text{并且}\quad p = \frac{1}{5}.$$ 在这些结果中，只有 $BBAAA$ 这一种情况下球队 B 赢得第一场比赛，因此其发生概率为 $1/5$。

Topics