If two subsets \( A \) and \( B \) are selected at random from a set \( S \) containing \( n \) elements, then the probability that \( A \cap B = \emptyset \) and \( A \cup B = S \) is:
Show Hint
- \( \frac{1}{2^n} \)
- \( 2^n \)
- \( \frac{1}{2^{n+1}} \)
- \( \frac{1}{2^n \times 2^n} \)
The Correct Option is A
Solution and Explanation
We are given that two subsets \( A \) and \( B \) are selected at random from a set \( S \) containing \( n \) elements, and we need to find the probability that \( A \cap B = \emptyset \) and \( A \cup B = S \). For each element in the set \( S \), there are three possibilities:
1. The element is only in \( A \).
2. The element is only in \( B \). 3. The element is in neither \( A \) nor \( B \). However, for the condition \( A \cap B = \emptyset \), an element cannot be in both \( A \) and \( B \) simultaneously. So, for each element, there are two choices: 1. The element is in \( A \). 2. The element is in \( B \). Now, for the condition \( A \cup B = S \), every element of \( S \) must be either in \( A \) or in \( B \) (but not both). Hence, there are \( 2^n \) possible ways to assign each of the \( n \) elements to either \( A \) or \( B \), and the total number of ways is \( 2^n \). The total number of ways to choose \( A \) and \( B \) from \( S \) without any restrictions is \( 3^n \), as each element can independently belong to \( A \), \( B \), or neither. Thus, the probability is the ratio of favorable outcomes to total outcomes: \[ \frac{2^n}{3^n} = \frac{1}{2^n} \] Thus, the correct answer is \( \frac{1}{2^n} \).
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