+27 Let \(S = \{1, 2, 3, \dots, n\}.\)Three subsets A, B, C of S are chosen at random.
(a) Find the probability that \(A \cup B \cup C = S.\)
(b) Find the probability that \(A \subseteq B \subseteq C.\)
Can anyone use complete sentences and explain all steps to the solution? Help would be appreciated!
+1768 Part (a): The number of ways to choose three subsets A, B, and C of S is 2^n * 2^n * 2^n = 2^(3n).
Now, let's count the number of ways to choose A, B, and C such that their union is S. We can do this by considering the complement of the union, which is the intersection of the complements of A, B, and C. This intersection must be the empty set.
There are 2^n ways to choose the complement of A, 2^n ways to choose the complement of B, and 2^n ways to choose the complement of C.
However, we have overcounted the number of ways to choose the empty set as the intersection, because there are 3 ways to choose which complement is empty and the other two complements are not.
Therefore, the number of ways to choose A, B, and C such that their union is S is 2^(3n) - 3 * 2^(2n).
Finally, the probability that the union of A, B, and C is S is the number of ways to choose A, B, and C such that their union is S divided by the total number of ways to choose A, B, and C. Therefore, the probability is:
(2^(3n) - 3 * 2^(2n)) / 2^(3n) = 1 - 3/2^n
As n approaches infinity, the probability approaches 1. This means that it is almost certain that the union of three randomly chosen subsets of a large set will be the entire set.
+1222 Here is part (b):
The number of ways to choose three subsets A, B, C of S is 2^n * 2^n * 2^n = 2^(3n).
To find the number of ways that A is a subset of B, and B is a subset of C, we can use the following approach:
Choose a subset C of S. There are 2^n ways to do this.
Choose a subset B of C. There are 2^|C| ways to do this, where |C| is the size of C.
Choose a subset A of B. There are 2^|B| ways to do this, where |B| is the size of B.
Therefore, the number of ways that A is a subset of B, and B is a subset of C is:
∑_{C⊆S} 2^|C| * 2^|B|
where the sum is over all subsets C of S.
We can simplify this expression by noting that for any fixed subset C of S, the number of ways to choose subsets B and A of C such that A is a subset of B is 2^(|C|+1). This is because we can choose any subset B of C, and then choose any subset A of B. There are 2^(|C|+1) ways to do this.
Therefore, the number of ways that A is a subset of B, and B is a subset of C is:
∑_{C⊆S} 2^(|C|+1)
We can evaluate this sum using the following trick:
∑_{C⊆S} 2^(|C|+1) = 2 * ∑_{C⊆S} 2^|C|
The sum on the right-hand side is just the number of subsets of S, which is 2^n. Therefore, we have:
∑_{C⊆S} 2^(|C|+1) = 2 * 2^n = 2^(n+1)
Finally, the probability that A is a subset of B, and B is a subset of C is:
2^(n+1) / 2^(3n) = 1 / 2^(2n-1)
Therefore, the probability that A is a subset of B, and B is a subset of C is 1 / 2^(2n-1).
