1986 AMC 12 第 22 题

Question

从

\{1,2,3,\ldots,10\}

中随机选取六个互不相同的整数。在所选整数中，第二小的数是

3

的概率是多少？

Six distinct integers are picked at random from

\{1,2,3,\ldots,10\}.

What is the probability that, among those selected, the second smallest is

3?

\frac1{60}

\frac16

\frac13

\frac12

以上均非

none of these

先数出这十个整数的所有六元子集

Count all six-element subsets of the ten integers

若第二小的数是

3

，应从小于

3

的数中选一个，从大于

3

的数中选四个

For second-smallest

3,

choose one element below

3

and four above

3

\binom{10}{6}=210

个可能的集合。若第二小的元素是

3

，则必须选取

3

，再从

\{1,2\}

中选一个，并从

\{4,5,\ldots,10\}

中选四个。因此满足条件的集合有

2\binom74=70

个。所求概率为

\frac{70}{210}=\frac{1}{3}

。

因此正确答案是 C。

There are

\binom{10}{6}=210

possible sets. If the second-smallest element is

3,

then

3

is selected, one element is chosen from

\{1,2\},

and four are chosen from

\{4,5,\ldots,10\}.

This gives

2\binom74=70

favorable sets. The probability is

\frac{70}{210}=\frac{1}{3}.

Thus the correct answer is C.

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