To solve the problem, we start by examining the total number of ways to choose subsets from the set \( S = {1, 2, 3, \ldots, n} \).
Number of Subsets: For each element in \( S \), we have two choices: either to include it in a subset or not to include it. Thus, for \( n \) elements, the total number of possible subsets (including the empty set) is \( 2^n \).
Since we are selecting three subsets \( A \), \( B \), and \( C \) independently, the total number of ways to choose the three subsets is \( (2^n)^3 = 2^{3n} \).
Now, let's answer the specific parts of the question:
(a) Find the probability that \( A \) contains all elements \( 1, 2, 3, \ldots, n \).
To satisfy the condition that \( A \) contains all the elements \( 1, 2, 3, \ldots, n \), the only choice for subset \( A \) is to include all \( n \) elements. Thus, there is exactly 1 way to choose \( A \) such that it contains all elements.
The number of ways to choose subsets \( B \) and \( C \) remains \( 2^n \) for each. Thus, the number of favorable outcomes for subsets \( A \), \( B \), and \( C \) is:
\[ 1 \cdot 2^n \cdot 2^n = 2^{2n}. \]
Therefore, the probability that \( A \) contains all elements is:
\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2^{2n}}{2^{3n}} = \frac{1}{2^n}. \]
(b) Find the probability that \( B \) contains all the even numbers.
The even numbers in set \( S \) are \( 2, 4, 6, \ldots, 2\lfloor \frac{n}{2} \rfloor \). The number of even numbers is \( \lfloor \frac{n}{2} \rfloor \). For subset \( B \) to include all even numbers, there is only 1 way to choose those.
The remaining elements of \( S \) are the odd numbers, and there are \( n - \lfloor \frac{n}{2} \rfloor = \lceil \frac{n}{2} \rceil \) odd numbers. Each of these odd numbers can either be in \( B \) or not, leading to \( 2^{\lceil n/2 \rceil} \) choices for the odd numbers.
The number of ways to choose subsets \( A \) and \( C \) remains \( 2^n \) for each. Therefore, the number of favorable outcomes is:
\[ 1 \cdot 2^n \cdot 2^{\lceil n/2 \rceil} = 2^{n + \lceil n/2 \rceil}. \]
Thus, the probability that \( B \) contains all even numbers is:
\[ \text{Probability} = \frac{2^{n + \lceil n/2 \rceil}}{2^{3n}} = \frac{2^{n + \lceil n/2 \rceil}}{2^{3n}} = \frac{2^{\lceil n/2 \rceil - 2n}}{1} = 2^{\lceil n/2 \rceil - 2n}. \]
(c) Find the probability that \( C \) contains all the odd numbers.
Using a similar reasoning as for subset \( B \), there are \( \lfloor \frac{n}{2} \rfloor \) odd numbers in \( S \). For \( C \) to contain all odd numbers, there is again only 1 way to choose those.
The remaining even numbers can be included or excluded freely. The number of ways to choose from the even numbers is \( 2^{\lfloor n/2 \rfloor} \).
The number of ways to choose subsets \( A \) and \( B \) is still \( 2^n \) for each. Therefore, the number of favorable outcomes is:
\[ 1 \cdot 2^n \cdot 2^{\lfloor n/2 \rfloor} = 2^{n + \lfloor n/2 \rfloor}. \]
Thus, the probability that \( C \) contains all odd numbers is:
\[ \text{Probability} = \frac{2^{n + \lfloor n/2 \rfloor}}{2^{3n}} = \frac{2^{n + \lfloor n/2 \rfloor}}{2^{3n}} = 2^{\lfloor n/2 \rfloor - 2n}. \]
Putting it all together, the answers to the parts are:
(a) \( \frac{1}{2^n} \)
(b) \( 2^{\lceil n/2 \rceil - 2n} \)
(c) \( 2^{\lfloor n/2 \rfloor - 2n} \)