Question

从 $1000$ 到 $9999$ （含端点）之间随机选一个整数。它是奇数且各位数字互不相同的概率是多少？

An integer between $1000$ and $9999,$ inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

$\dfrac{14}{75}$

$\dfrac{56}{225}$

$\dfrac{107}{400}$

$\dfrac{7}{25}$

$\dfrac{9}{25}$

答案：B

知识点：基本概率乘法原理

难度评级：1550

视频讲解：

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文字解答：

这个数是奇数，所以个位数字有 $5$ 种选择。千位数字不能为零，也不能等于个位数字，因此有 $8$ 种选择。百位有 $8$ 种选择，十位有 $7$ 种选择。符合条件的数共有 $5 \cdot 8 \cdot 8 \cdot 7 = 2240$ ，所以概率为 $\dfrac{2240}{9000} = \dfrac{56}{225}$ 。

所以正确答案是 B。

Since the number is odd, the last digit is odd, giving $5$ possibilities. The thousands digit cannot be zero or the number we already got, so that gives $8$ possibilities. Similarly, the hundreds digit has $8$ possibilities, and the tens digit has $7$ possibilities. This gives a total of $5 \cdot 8 \cdot 8 \cdot 7 = 2240,$ making the probability $\dfrac{2240}{9000} = \dfrac{56}{225}.$

Thus, B is the correct answer.

2017 AMC 8 第 20 题

答案：B

难度评级：1550

视频讲解：

文字解答：

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