位于 1 到 100(包括两者)之间的给定正整数不能被 2、3 或 5 整除的概率是______。
(一) 0.259
(乙) 0.459
(C) 0.325
(D) 0.225答案：(一)
解释：
There are total 100 numbers, out of which
50 numbers are divisible by 2,
33 numbers are divisible by 3,
20 numbers are divisible by 5
Following are counted twice above
16 numbers are divisible by both 2 and 3
10 numbers are divisible by both 2 and 5
6 numbers are divisible by both 3 and 5
Following is counted thrice above
3 numbers are divisible by all 2, 3 and 5
所以可被 2、3 和 5 整除的总数是 =
= 50 + 33 + 20 – 16 – 10 – 6 + 3
= 103 – 29
= 74
所以数字是数字的概率不是
可被 2、3 和 5 整除 = (100 – 74)/100 = 0.26
这个问题的测验