To find the probability that the subset \( A \) is a subset of \( B \), and \( B \) is a subset of \( C \) for randomly chosen subsets \( A, B, C \) of the set \( S = { 1, 2, 3, \ldots, n } \), we can analyze the choices we have for each element in \( S \).
Each element of \( S \) can be in three states with respect to the subsets \( A, B, C \):
- The element is not in \( C \).
- The element is in \( C \) but not in \( B \).
- The element is in \( B \) but not in \( A \).
- The element is in \( A \) (which also means it must be in \( B \) and \( C \) based on the subset conditions).
Since \( A \subseteq B \subseteq C \) must hold, we can characterize our decision for each element:
- If the element is not in \( C \), then it cannot be in \( B \) or \( A \).
- If the element is in \( C \), it can either be:
- Not in \( B \) (and not in \( A \)), or
- In \( B \) (and therefore also in \( A \)).
Let’s summarize the choices for each element:
- Not in \( C \): In this case, it is guaranteed not to be in \( B \) and thus not in \( A \).
- In \( C \) and not in \( B \): It is not in \( A \).
- In \( C \) and in \( B \): In this case, it could be either in \( A \) or not.
- If it’s in \( A \), then it is also in \( B \) and \( C \).
- If it’s not in \( A \), then it’s still in \( C \) and \( B \).
So, we can break down the choices into the following:
- For each of the \( n \) elements, we have 3 choices based on our earlier classification:
- The element is not in \( C \) (and therefore not in \( B \) or \( A \)).
- The element is in \( C \) but not in \( B \) (and therefore not in \( A \)).
- The element is in \( C \) and \( B \) (from which we can choose to include it in \( A \) or not).
Thus:
- For an element that is included in both \( C \) and \( B \), we have 2 choices (to be in \( A \) or not),
- For elements not in \( C \), we only have one choice.
Hence, for every element in \( S \), the choices can be calculated as follows:
- \( 1 \) way not to include it in \( C \),
- \( 1 \) way to include it in \( C \) and not in \( B \),
- \( 2 \) ways to include it in \( C\) and \( B \) (for it being either in \( A \) or not).
This gives us a total of \( 1 + 1 + 2 = 4 \) choices for each element.
Since there are \( n \) elements in \( S \), the total number of combinations for choosing subsets \( A \), \( B \), and \( C \) that satisfy \( A \subseteq B \subseteq C \) is \( 4^n \).
On the other hand, the total number of ways to choose arbitrary subsets \( A \), \( B \), and \( C \) independently from \( S \) is \( 2^n \times 2^n \times 2^n = (2^n)^3 = 8^n \).
Thus, the probability \( P \) that \( A \) is a subset of \( B \) and \( B \) is a subset of \( C \) is given by
\[ P = \frac{4^n}{8^n} = \left(\frac{4}{8}\right)^n = \left(\frac{1}{2}\right)^n. \]
Therefore, the final answer is
\[ \boxed{\left( \frac{1}{2} \right)^n}. \]
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