AMC10 2016 A

AMC10 2016 A · Q12

AMC10 2016 A · Q12. It mainly tests Probability (basic), Parity (odd/even).

Three distinct integers are selected at random between 1 and 2016, inclusive. Which of the following is a correct statement about the probability $p$ that the product of the three integers is odd?

从 1 到 2016（含）之间随机选取三个互不相同的整数。以下哪一项关于这三个整数的乘积为奇数的概率 $p$ 的说法是正确的？

(A) $p < \frac{1}{8}$ $p < \frac{1}{8}$

(B) $p = \frac{1}{8}$ $p = \frac{1}{8}$

(D) $p = \frac{1}{3}$ $p = \frac{1}{3}$

(E) $p > \frac{1}{3}$ $p > \frac{1}{3}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The product of three integers is odd if and only if all three integers are odd. There are 1008 odd integers among the 2016 integers in the given range. The probability that all the selected integers are odd is $$ p=\frac{1008}{2016}\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}. $$ The first factor is $\frac{1}{2}$ and each of the other factors is less than $\frac{1}{2}$, so $p<\frac{1}{8}$.

答案（A）：三个整数的乘积为奇数当且仅当这三个整数都为奇数。在给定范围内的 2016 个整数中，有 1008 个奇数。所选整数全为奇数的概率为 $$ p=\frac{1008}{2016}\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}. $$ 第一个因子是 $\frac{1}{2}$，其余每个因子都小于 $\frac{1}{2}$，因此 $p<\frac{1}{8}$。

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