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 \( \frac{1}{2^n} \).
Three distinct numbers are selected randomly from the set \( \{1, 2, 3, \dots, 40\} \). If the probability, that the selected numbers are in an increasing G.P. is \( \frac{m}{n} \), where \( \gcd(m, n) = 1 \), then \( m + n \) is equal to:
A board has 16 squares as shown in the figure. Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is: