AMC12 2017 B

AMC12 2017 B · Q22

AMC12 2017 B · Q22. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn—one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

Abby、Bernardo、Carl和Debra玩一个游戏，每人起始有4个硬币。游戏有4轮。每轮中，将4个球放入一个瓮中——一个绿色、一个红色、两个白色。玩家依次无放回抽取球。抽到绿色球的人给抽到红色球的人一个硬币。第四轮结束后，每位玩家都有4个硬币的概率是多少？

(A) \frac{7}{576} \frac{7}{576}

(B) \frac{5}{192} \frac{5}{192}

(D) \frac{5}{144} \frac{5}{144}

(E) \frac{7}{48} \frac{7}{48}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): There are $4\cdot 3=12$ outcomes for each set of draws and therefore $12^4$ outcomes in all. To count the number of outcomes in which each player will end up with four coins, note that this can happen in four ways: • For some permutation $(w,x,y,z)$ of $\{\text{Abby},\text{Bernardo},\text{Carl},\text{Debra}\}$, the outcomes of the four draws are that $w$ gives a coin to $x$, $x$ gives a coin to $y$, $y$ gives a coin to $z$, and $z$ gives a coin to $w$, in one of $4!=24$ orders. There are $3$ ways to choose whom Abby gives her coin to and $2$ ways to choose whom that person gives his or her coin to, which makes $6$ ways to choose the givers and receivers for these transaction. Therefore there are $24\cdot 6=144$ ways for this to happen. • One pair of the players exchange coins, and the other two players also exchange coins, in one of $4!=24$ orders. There are $3$ ways to choose the pairings. Therefore there are $24\cdot 3=72$ ways for this to happen. • Two of the players exchange coins twice. There are $\binom{4}{2}=6$ ways to choose those players and $\binom{4}{2}=6$ ways to choose the orders of the exchanges, for a total of $6\cdot 6=36$ ways for this to happen. • One of the players is involved in all four transactions, giving and receiving a coin from each of two others. There are $4$ ways to choose this player, $3$ ways to choose the other two players, and $4!=24$ ways to choose the order in which the transactions will take place. Therefore there are $4\cdot 3\cdot 24=288$ ways for this to happen. In all, there are $144+72+36+288=540$ outcomes that will result in each player having four coins. The requested probability is $\dfrac{540}{12^4}=\dfrac{5}{192}$.

答案（B）：每一轮抽取有 $4\cdot 3=12$ 种结果，因此总共有 $12^4$ 种结果。要计算使得每个玩家最终都有四枚硬币的结果数，注意这种情况有四种方式： • 对于 $\{\text{Abby},\text{Bernardo},\text{Carl},\text{Debra}\}$ 的某个排列 $(w,x,y,z)$，四次抽取的结果是：$w$ 给 $x$ 一枚硬币，$x$ 给 $y$ 一枚硬币，$y$ 给 $z$ 一枚硬币，$z$ 给 $w$ 一枚硬币；其发生顺序有 $4!=24$ 种。Abby 把硬币给谁有 $3$ 种选法，而该人再把硬币给谁有 $2$ 种选法，因此确定这些交易的给出者与接收者共有 $6$ 种方式。故这种情形共有 $24\cdot 6=144$ 种。 • 一对玩家互换硬币，另外两位玩家也互换硬币；发生顺序有 $4!=24$ 种。配对方式有 $3$ 种，因此共有 $24\cdot 3=72$ 种。 • 两位玩家互换硬币两次。选择这两位玩家有 $\binom{4}{2}=6$ 种，选择两次交换的先后顺序也有 $\binom{4}{2}=6$ 种，因此共有 $6\cdot 6=36$ 种。 • 有一位玩家参与全部四次交易，并分别与另外两位玩家各进行一次“给”和一次“收”。选择该玩家有 $4$ 种，选择另外两位玩家有 $3$ 种，交易发生顺序有 $4!=24$ 种，因此共有 $4\cdot 3\cdot 24=288$ 种。总计有 $144+72+36+288=540$ 种结果会使得每位玩家都有四枚硬币。所求概率为 $\dfrac{540}{12^4}=\dfrac{5}{192}$。

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