A subset A of the set X={1,2,3,..,100} is chosen at random. The set X is reconstructed by replacing the elements of A, and another subset B of X is chosen at random. The probability that `A cap B` contains exactly 10 elements, is

Options

A. `.^(100)C_(10)((3)/(4))^(90)`
B. `.^(100)C_(10)((1)/(2))^(100)`
C. `.^(100)C_(10)xx(3^(90))/(4^(100))`
D. none of these

Answer

To solve the problem, we need to find the probability that the intersection of two randomly chosen subsets \( A \) and \( B \) from the set \( X = \{1, 2, 3, \ldots, 100\} \) contains exactly 10 elements.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a set \( X \) with 100 elements. We are choosing two subsets \( A \) and \( B \) randomly from \( X \). We need to calculate the probability that the intersection \( A \cap B \) has exactly 10 elements.

2. **Total Possible Subsets**:
The total number of subsets of a set with \( n \) elements is \( 2^n \). Therefore, the total number of subsets for \( X \) (which has 100 elements) is:
\[ 2^{100} \]
Since both \( A \) and \( B \) can be chosen independently, the total number of ways to choose both subsets \( A \) and \( B \) is:
\[ 2^{100} \times 2^{100} = 4^{100} \]

3. **Choosing Elements for the Intersection**:
We want \( A \cap B \) to contain exactly 10 elements. We can choose 10 elements from the 100 elements in \( X \). The number of ways to choose 10 elements from 100 is given by the binomial coefficient:
\[ \binom{100}{10} \]

4. **Choosing Remaining Elements**:
After choosing 10 elements for the intersection, we have 90 elements left in \( X \). Each of these remaining 90 elements can either be included in \( A \) or not included in \( A \) (2 choices), and can also either be included in \( B \) or not included in \( B \) (2 choices). Thus, for each of the 90 elements, there are 4 possible combinations (included in both, included in \( A \) only, included in \( B \) only, or included in neither). Therefore, the number of ways to choose the remaining elements is:
\[ 4^{90} \]

5. **Calculating the Favorable Outcomes**:
The total number of favorable outcomes where \( A \cap B \) contains exactly 10 elements is given by:
\[ \binom{100}{10} \times 4^{90} \]

6. **Calculating the Probability**:
The probability \( P \) that \( A \cap B \) contains exactly 10 elements is the ratio of the number of favorable outcomes to the total number of outcomes:
\[ P = \frac{\binom{100}{10} \times 4^{90}}{4^{100}} = \frac{\binom{100}{10}}{4^{10}} \]

### Final Answer:
The probability that \( A \cap B \) contains exactly 10 elements is:
\[ P = \frac{\binom{100}{10}}{4^{10}} \]