AMC12 2019 A

AMC12 2019 A · Q20

AMC12 2019 A · Q20. It mainly tests Probability (basic), Conditional probability (basic).

Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0, 1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x - y| > \frac{1}{2}$?

实数在0到1之间（包含端点）以以下方式选择。抛一次公平硬币。如果正面，则再抛一次，第二抛为正面则选0，反面则选1。另一方面，如果第一抛为反面，则从闭区间$[0, 1]$中均匀随机选择该数。独立选择两个随机数$x$和$y$。求$|x - y| > \frac{1}{2}$的概率？

(A) \frac{1}{3} \frac{1}{3}

(B) \frac{7}{16} \frac{7}{16}

(D) \frac{9}{16} \frac{9}{16}

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

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The probability that the first coin flip for both $x$ and $y$ is heads is $\frac{1}{4}$, and in half of these cases $|x-y|$ will be $0$ and in the other half of these cases $|x-y|$ will be $1$. This contributes $\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$ to the probability that $|x-y|>\frac{1}{2}$. The probability that the first coin flip for $x$ is heads and the first coin flip for $y$ is tails or vice versa is $\frac{1}{2}$. In such cases, one of the variables is $0$ or $1$, and the probability that $|x-y|>\frac{1}{2}$ is $\frac{1}{2}$. This contributes $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$ to the probability that $|x-y|>\frac{1}{2}$. Finally, $\frac{1}{4}$ of the time both $x$ and $y$ will be chosen uniformly from $[0,1]$. In this case, the situation can be modeled by the following diagram, in which the area of the shaded region gives the probability that $|x-y|>\frac{1}{2}$. This contributes $\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}$ to the probability that $|x-y|>\frac{1}{2}$. The requested probability is $\frac{1}{8}+\frac{1}{4}+\frac{1}{16}=\frac{7}{16}$.

答案（B）：$x$ 和 $y$ 的第一次抛硬币都为正面的概率是 $\frac{1}{4}$。在这些情况的一半中，$|x-y|$ 会是 $0$，另一半中 $|x-y|$ 会是 $1$。因此对事件 $|x-y|>\frac{1}{2}$ 的概率贡献为 $\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$。 $x$ 第一次为正面且 $y$ 第一次为反面（或相反）的概率是 $\frac{1}{2}$。在这种情况下，其中一个变量为 $0$ 或 $1$，而 $|x-y|>\frac{1}{2}$ 的概率为 $\frac{1}{2}$。因此对事件 $|x-y|>\frac{1}{2}$ 的概率贡献为 $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$。最后，有 $\frac{1}{4}$ 的时间 $x$ 和 $y$ 都从区间 $[0,1]$ 上服从均匀分布选取。在这种情况下，可以用下图来建模，其中阴影区域的面积给出 $|x-y|>\frac{1}{2}$ 的概率。于是对事件 $|x-y|>\frac{1}{2}$ 的概率贡献为 $\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}$。所求概率为 $\frac{1}{8}+\frac{1}{4}+\frac{1}{16}=\frac{7}{16}$。

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