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 \( \frac{1}{2^n} \).
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :