Question

一枚硬币有偏，且每次抛掷出现正面的概率为 $\dfrac{2}{3}$ ，出现反面的概率为 $\dfrac{1}{3}$ 。各次抛掷结果相互独立。一名玩家可以选择玩游戏 A 或游戏 B。在游戏 A 中，她抛掷硬币三次，若三次结果全相同则获胜。在游戏 B 中，她抛掷硬币四次，若第一次和第二次的结果相同且第三次和第四次的结果相同，则获胜。游戏 A 的获胜机会与游戏 B 的获胜机会相比如何？

A coin is biased in such a way that on each toss the probability of heads is $\dfrac{2}{3}$ and the probability of tails is $\dfrac{1}{3}.$ The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?

游戏 A 的获胜概率比游戏 B 的获胜概率小 $\dfrac{4}{81}$ 。

The probability of winning Game A is $\dfrac{4}{81}$ less than the probability of winning Game B.

游戏 A 的获胜概率比游戏 B 的获胜概率小 $\dfrac{2}{81}$ 。

The probability of winning Game A is $\dfrac{2}{81}$ less than the probability of winning Game B.

两个概率相同。

The probabilities are the same.

游戏 A 的获胜概率比游戏 B 的获胜概率大 $\dfrac{2}{81}$ 。

The probability of winning Game A is $\dfrac{2}{81}$ greater than the probability of winning Game B.

游戏 A 的获胜概率比游戏 B 的获胜概率大 $\dfrac{4}{81}$ 。

The probability of winning Game A is $\dfrac{4}{81}$ greater than the probability of winning Game B.

答案：D

知识点：独立事件基本概率

难度评级：1800

解答：

令 $p = \dfrac23$ 。游戏 A 在三次抛掷结果全部相同时获胜，概率为 $p^3 + (1-p)^3$ 。游戏 B 要求第一对相同且第二对相同，每对相同的概率为 $p^2 + (1-p)^2$ ，所以获胜概率为 $\bigl(p^2 + (1-p)^2\bigr)^2$ 。当 $p = \tfrac23$ 时，游戏 A 的概率为 $\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13$ ，游戏 B 的概率为 $\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}$ 。差为 $\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81}$ ，所以游戏 A 的胜率高 $\tfrac{2}{81}$ 。

所以正确答案是 D。

Let $p = \dfrac23.$ Game A is won when all three tosses match: $p^3 + (1-p)^3.$ Game B needs the first pair to match and the second pair to match, each with probability $p^2 + (1-p)^2,$ so the win probability is $\bigl(p^2 + (1-p)^2\bigr)^2.$ With $p = \tfrac23,$ Game A gives $\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13,$ and Game B gives $\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}.$ The difference is $\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81},$ so Game A is $\tfrac{2}{81}$ more likely.

Thus, the correct answer is D.

2017 AMC 12B 第 17 题

答案：D

难度评级：1800

解答：

其他年份的第 17 题