Question

从 $0$ 到 $1$ 之间独立均匀随机地选出三个实数。这三个数中最大的那个数大于另外两个数各自的 $2$ 倍的概率是多少？（换句话说，如果所选的数为 $a \ge b \ge c$ ，则 $a \gt 2b$ 。）

Three real numbers are chosen independently and uniformly at random between $0$ and $1.$ What is the probability that the greatest of these three numbers is greater than $2$ times each of the other two numbers? (In other words, if the chosen numbers are $a \ge b \ge c,$ then $a \gt 2b.$ )

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

答案：E

知识点：几何概率微积分

难度评级：2270

解答：

将数按大小排列为 $x_1 \gt x_2 \gt x_3$ ；顺序统计量在这个区域上的联合密度为 $6$ 。所求事件为 $x_1 \gt 2x_2$ 。

对 $x_3$ 从 $0$ 到 $x_2$ 积分，会贡献因子 $x_2$ 。于是 $\begin{aligned} P &= 6\int_0^{1/2} x_2\int_{2x_2}^{1} dx_1\, dx_2 \\ &= 6\int_0^{1/2} x_2(1 - 2x_2)\, dx_2. \end{aligned}$

这等于 $6\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}$ 。

因此，正确答案是 E。

Order the values as $x_1 \gt x_2 \gt x_3$ ; the joint density of the order statistics is $6$ on this region. The event is $x_1 \gt 2x_2.$

Integrating $x_3$ from $0$ to $x_2$ contributes a factor of $x_2.$ Then $\begin{aligned} P &= 6\int_0^{1/2} x_2\int_{2x_2}^{1} dx_1\, dx_2 \\ &= 6\int_0^{1/2} x_2(1 - 2x_2)\, dx_2. \end{aligned}$

This equals $6\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}.$

Thus, the correct answer is E.

2025 AMC 12A 第 22 题

答案：E

难度评级：2270

解答：

其他年份的第 22 题