AMC12 2017 B

AMC12 2017 B · Q17

AMC12 2017 B · Q17. It mainly tests Probability (basic), Conditional probability (basic).

A coin is biased in such a way that on each toss the probability of heads is $\frac{2}{3}$ and the probability of tails is $\frac{1}{3}$. 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?

一枚硬币是偏倚的，每次抛掷正面概率为 $\frac{2}{3}$，反面概率为 $\frac{1}{3}$。抛掷结果相互独立。玩家可以选择玩游戏 A 或游戏 B。在游戏 A 中，她抛三次硬币，如果三次结果都相同则获胜。在游戏 B 中，她抛四次硬币，如果第一次和第二次结果相同且第三次和第四次结果相同则获胜。游戏 A 的获胜机会与游戏 B 的获胜机会相比如何？

(A) The probability of winning Game A is $\frac{4}{81}$ less than the probability of winning Game B. 赢得 A 的概率比赢得 B 少 $\frac{4}{81}$。

(B) The probability of winning Game A is $\frac{2}{81}$ less than the probability of winning Game B. 赢得 A 的概率比赢得 B 少 $\frac{2}{81}$。

(D) The probability of winning Game A is $\frac{2}{81}$ greater than the probability of winning Game B. 赢得 A 的概率比赢得 B 多 $\frac{2}{81}$。

(E) The probability of winning Game A is $\frac{4}{81}$ greater than the probability of winning Game B. 赢得 A 的概率比赢得 B 多 $\frac{4}{81}$。

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $p$ be the probability of heads. To win Game A requires that all three tosses be heads, which occurs with probability $p^3$, or all three tosses be tails, which occurs with probability $(1-p)^3$. To win Game B requires that the first two tosses be the same, the probability of which is $p^2+(1-p)^2$, and that the last two tosses be the same, which occurs with the same probability. Therefore the probability of winning Game A minus the probability of winning Game B is $$(p^3+(1-p)^3)-(p^2+(1-p)^2)^2.$$ As $p=\frac{2}{3}$, this gives $$\left(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3\right)-\left(\left(\frac{2}{3}\right)^2+\left(\frac{1}{3}\right)^2\right)^2=\frac{1}{3}-\frac{25}{81}=\frac{2}{81}.$$ Thus the probability of winning Game A is $\frac{2}{81}$ greater than the probability of winning Game B.

答案（D）：设 $p$ 为正面朝上的概率。要赢游戏 A，需要三次投掷都为正面，其概率为 $p^3$；或者三次投掷都为反面，其概率为 $(1-p)^3$。要赢游戏 B，需要前两次投掷结果相同，其概率为 $p^2+(1-p)^2$；并且后两次投掷结果也相同，而这同样以相同的概率发生。因此，赢游戏 A 的概率减去赢游戏 B 的概率为 $$(p^3+(1-p)^3)-(p^2+(1-p)^2)^2.$$ 当 $p=\frac{2}{3}$ 时， $$\left(\left(\frac{2}{3}\right)^3+\left(\frac{1}{3}\right)^3\right)-\left(\left(\frac{2}{3}\right)^2+\left(\frac{1}{3}\right)^2\right)^2=\frac{1}{3}-\frac{25}{81}=\frac{2}{81}.$$ 因此，赢游戏 A 的概率比赢游戏 B 的概率大 $\frac{2}{81}$。

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