2007 AMC 10A 第 16 题

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16.

整数 a,b,ca, b, cdd 不要求互不相同,分别独立地从 0020072007(含端点)中随机选择。adbcad - bc 为偶数的概率是多少?

Integers a,b,c,a, b, c, and d,d, not necessarily distinct, are chosen independently and at random from 00 to 2007,2007, inclusive. What is the probability that adbcad - bc is even?

38\dfrac{3}{8}

716\dfrac{7}{16}

12\dfrac{1}{2}

916\dfrac{9}{16}

58\dfrac{5}{8}

答案:E
知识点:奇偶性基本概率独立事件
难度评级:1540
解答:

0020072007 的整数中一半为奇数,所以 adad 为奇数的概率为 1212=14\tfrac12 \cdot \tfrac12 = \tfrac14,为偶数的概率为 34\tfrac34bcbc 也是如此。

adbcad - bc 为偶数当两个乘积奇偶性相同: 1414+3434=116+916=58. \tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58.

所以正确答案是 E

Half the integers from 00 to 20072007 are odd, so each of adad and bcbc is odd with probability 1212=14\tfrac12 \cdot \tfrac12 = \tfrac14 and even with probability 34.\tfrac34.

The difference adbcad - bc is even when both products have the same parity: 1414+3434=116+916=58. \tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58.

Thus, the correct answer is E.

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