A class of problems that requires one to use permutations and combinations for computing probability has at its heart notion of sets and subsets. They are generally an abstract formulation of some concrete situation and require the application of counting techniques.
A is a set containing 10 elements. A subset `P_(1)` of A is chosen and the set A is chosen and the set A is reconstructed by replacing the elements of `P_(1)`. A subset `P_(2)` of A is chosen and again the set A is reconstructed by replacing the elements of `P_(2)`. This process is continued by choosing subsets `P_(1), P_(2), ... P_(10)`.
The probability that `P_(1)capP_(2)cap...P_(10)=phi` is

Options

A. A ) `(1023)^(10)/(2^(100))`
B. B) `(1025)^(10)/(4^(100))`
C. C) `(1023)^(10)/(4^(100))`
D. D) `(1025)^(10)/(2^(100))`

Answer

To solve the problem, we need to find the probability that the intersection of the subsets \( P_1, P_2, \ldots, P_{10} \) is empty, i.e., \( P_1 \cap P_2 \cap \ldots \cap P_{10} = \emptyset \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a set \( A \) containing 10 elements. We are selecting subsets \( P_1, P_2, \ldots, P_{10} \) from \( A \) and we want to find the probability that none of the subsets share any elements.

2. **Choosing Subsets**:
For each element in set \( A \), it can either be included in a subset \( P_k \) or not included. Thus, for each element, there are 2 choices (to be in \( P_k \) or not).

3. **Total Choices for Each Element**:
Since there are 10 elements in set \( A \), the total number of ways to choose subsets \( P_1, P_2, \ldots, P_{10} \) is given by:
\[ 2^{10} = 1024 \]

4. **Condition for Empty Intersection**:
For the intersection \( P_1 \cap P_2 \cap \ldots \cap P_{10} \) to be empty, each element of \( A \) must not belong to at least one of the subsets \( P_k \).

5. **Calculating Valid Cases**:
For each element \( a_i \) in \( A \):
- The element can either be included in none of the subsets or included in at least one subset.
- If \( a_i \) is included in none of the subsets, there is 1 way.
- If \( a_i \) is included in at least one subset, there are \( 2^{10} - 1 \) ways (since we exclude the case where it is in none).

Therefore, the number of ways for each element \( a_i \) to not belong to at least one subset is:
\[ 2^{10} - 1 \]

6. **Total Valid Cases for All Elements**:
Since there are 10 elements, the total number of ways in which none of the subsets share any elements is:
\[ (2^{10} - 1)^{10} \]

7. **Calculating the Probability**:
The probability \( P \) that \( P_1 \cap P_2 \cap \ldots \cap P_{10} = \emptyset \) is given by the ratio of the number of valid cases to the total cases:
\[ P = \frac{(2^{10} - 1)^{10}}{2^{10 \times 10}} = \frac{(1023)^{10}}{1024^{10}} = \left( \frac{1023}{1024} \right)^{10} \]

### Final Answer:
Thus, the probability that \( P_1 \cap P_2 \cap \ldots \cap P_{10} = \emptyset \) is:
\[ \left( \frac{1023}{1024} \right)^{10} \]