AMC8 2017

AMC8 2017 · Q20

AMC8 2017 · Q20. It mainly tests Basic counting (rules of product/sum), Probability (basic).

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?

随机选取一个1000到9999（包含端点）的整数。求其为奇数且各位数字均不同的概率？

(A) $\dfrac{14}{75}$ $\dfrac{14}{75}$

(B) $\dfrac{56}{225}$ $\dfrac{56}{225}$

(D) $\dfrac{7}{25}$ $\dfrac{7}{25}$

(E) $\dfrac{9}{25}$ $\dfrac{9}{25}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): There are 9000 integers between 1000 and 9999 inclusive. For an integer to be odd it must end in 1, 3, 5, 7, or 9. So there are 5 choices for the units digit. For a number to be between 1000 and 9999 the thousands digit must be nonzero and so there are now 8 choices for the thousands digit. For the hundreds digit there are 8 choices and for the tens digit there are 7 choices for a total number of $5 \cdot 8 \cdot 8 \cdot 7 = 2240$ choices. So the probability is \[\frac{2240}{9000}=\frac{56}{225}.\]

答案 (B)：在 1000 到 9999（含）之间共有 9000 个整数。一个整数要为奇数，其末位必须是 1、3、5、7 或 9，所以个位有 5 种选择。要使一个数介于 1000 和 9999 之间，千位必须非零，因此千位有 8 种选择。百位有 8 种选择，十位有 7 种选择，总共有 $5 \cdot 8 \cdot 8 \cdot 7 = 2240$ 个符合条件的数。因此所求概率为 \[\frac{2240}{9000}=\frac{56}{225}.\]

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