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随机子集

Random Subsets

专题
Probability / 概率
难度
L6

题目详情

从一个基数为 55 的集合 XX 的所有子集中,均匀随机地选取子集 AABB。求 AABB 的子集的概率。

Subsets AA and BB are chosen uniformly at random from the collections of all subsets of a set XX of cardinality 55. What is the probability that AA is a subset of BB?

解析

有序对子 (A,B)(A, B) 的总数为 2525=210=10242^5 * 2^5 = 2^{10} = 1024

对于每个元素 xx,若要求 ABA ⊆ B,则它在 (A,B)(A, B) 中的出现方式有 3 种合法选择。因此,有利的有序对子总数为 35=2433^5 = 243

进一步解释一下上面的说法。对每个元素 xx,它要么属于某个子集,要么不属于某个子集。因此对 AABB 来说:

  • xAx \in AxAx \notin A
  • xBx \in BxBx \notin B

对单个元素而言,AA 有两种选择,BB 也有两种选择,因此一共是 4 种可能。可以表示为有序对:

(xA?,xB?){(0,0)(0,1),(1,0),(1,1)}(x \in A?, x\in B?) \in \{(0,0)\, (0,1), (1,0), (1,1)\}

记住:ABA ⊆ B 的意思是,如果 xxAA 中,那么它也必须在 BB 中。

  • 情况 (0,0)(0, 0)xx 不在 AA 中,也不在 BB 中。允许。
  • 情况 (0,1)(0, 1)xx 不在 AA 中,但在 BB 中。允许。
  • 情况 (1,0)(1, 0)xxAA 中,但不在 BB 中。不允许。
  • 情况 (1,1)(1, 1)xx 同时在 AABB 中。允许。

因此,在 4 种等可能模式中,有 3 种满足 ABA ⊆ B

所以,AABB 的子集的概率为 2431024\frac{243}{1024}


Original Explanation

The total number of ordered pairs (A,B)(A, B) is 2525=210=10242^5 * 2^5 = 2^{10} = 1024.

For each element xx there are 3 valid choices for how it can appear in (A,B)(A, B) if we require ABA ⊆ B. So the total number of favorable ordered pairs is 35=2433^5 = 243.

To expand on the previous statement, we can think of each element xx as either in or not in a subset. So, for subsets AA and BB:

  • xAx \in A or xAx \notin A
  • xBx \in B or xBx \notin B

There's two choices for AA and two choices for BB which equals 4 total possibilities for this single element. We can represent them as ordered pairs:

(xA?,xB?){(0,0)(0,1),(1,0),(1,1)}(x \in A?, x\in B?) \in \{(0,0)\, (0,1), (1,0), (1,1)\}

Remember: ABA ⊆ B means if x is in AA, then it must also be in BB.

  • Case (0,0)(0, 0): x is neither in AA nor BB. Allowed (no conflict)
  • Case (0,1)(0, 1): x is not in AA but is in BB. Allowed (subset rule is still satisfied)
  • Case (1,0)(1, 0): x is in AA but not in BB. Not allowed
  • Case (1,1)(1, 1): x is in AA and BB: Allowed

So, out of the 4 equally likely patterns, 3 are valid when we require ABA ⊆ B.

Thus the probability that AA is a subset of BB is 2431024\frac{243}{1024}.