Question

按如下方式从 $0$ 到 $1$ 之间（含端点）选择一个实数。先抛一枚公平硬币。如果正面朝上，则再抛一次；第二次若正面朝上就选 $0$ ，若反面朝上就选 $1$ 。如果第一次抛硬币反面朝上，则从闭区间 $[0, 1]$ 中均匀随机选一个数。独立地用这种方式选出两个随机数 $x$ 和 $y$ 。求 $|x - y| \gt \dfrac{1}{2}$ 的概率。

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 $|x - y| \gt \dfrac{1}{2}?$

$\dfrac{1}{3}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{2}{3}$

答案：B

知识点：几何概率分类讨论

难度评级：2070

解答：

每个变量以概率 $\tfrac14$ 等于 $0$ ，以概率 $\tfrac14$ 等于 $1$ ，并以概率 $\tfrac12$ 在 $[0, 1]$ 上均匀分布。

考虑九种类型组合：数对 $(0, 1)$ 和 $(1, 0)$ 各贡献 $\tfrac{1}{16}$ 。四种“定点对均匀”的情况各贡献 $\tfrac{1}{16}$ 。 “均匀对均匀”的情况贡献 $\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}$ 。

总和为 $\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}$ 。

所以正确答案是 B。

Each variable equals $0$ with probability $\tfrac14,$ equals $1$ with probability $\tfrac14,$ and is uniform on $[0, 1]$ with probability $\tfrac12.$

Considering the nine combinations of types: the pairs $(0, 1)$ and $(1, 0)$ each contribute $\tfrac{1}{16}.$ Each of the four point-versus-uniform cases contributes $\tfrac{1}{16}.$ The uniform-versus-uniform case contributes $\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}.$

The total is $\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}.$

Thus, the correct answer is B.

2019 AMC 12A 第 20 题

答案：B

难度评级：2070

解答：

其他年份的第 20 题