Key Concepts and Formulas

• Probability Definition: The probability of an event $E$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes: $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
• Subset Formation and Element Placement: For a set $S$ with $n$ elements, there are $2^n$ possible subsets. When considering two subsets, say $A$ and $B$, from $S$, each element $x \in S$ can fall into one of four mutually exclusive regions:
  1. $x \in A$ and $x \in B$ (i.e., $x \in A \cap B$)
  2. $x \in A$ and $x \notin B$ (i.e., $x \in A \setminus B$)
  3. $x \notin A$ and $x \in B$ (i.e., $x \in B \setminus A$)
  4. $x \notin A$ and $x \notin B$ (i.e., $x \in S \setminus (A \cup B)$) This implies that for each element, there are 4 independent choices regarding its membership in the pair of subsets $(A, B)$.
• Combinations: The number of ways to choose $k$ elements from a set of $n$ distinct elements, without regard to the order of selection, is given by the combination formula: $^n C_k = \frac{n!}{k!(n-k)!}$

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Step-by-Step Solution

Step 1: Understand the Problem and Identify the Universal Set The given set is $S = \{1, 2, 3, 4, 5\}$, which has $n(S) = 5$ elements. We are randomly selecting two subsets, let's call them $A$ and $B$, from $S$. We need to find the probability that their intersection, $A \cap B$, contains exactly two elements.

Step 2: Calculate the Total Number of Possible Ways to Select Two Subsets We need to determine the total number of distinct ordered pairs of subsets $(A, B)$ that can be formed from the set $S$. As per the Key Concept of element placement, for each of the 5 elements in $S$, there are 4 independent choices regarding its membership in the subsets $A$ and $B$.

• For element 1, there are 4 choices.
• For element 2, there are 4 choices.
• ...
• For element 5, there are 4 choices. Since there are 5 elements, the total number of ways to define the two subsets $A$ and $B$ (i.e., the total number of ordered pairs of subsets) is: $\text{Total Number of Outcomes} = 4^5$ We can also write $4^5$ as $(2^2)^5 = 2^{10}$. $\text{Total Number of Outcomes} = 2^{10} = 1024$

Step 3: Calculate the Number of Favorable Outcomes (Subsets with Exactly Two Elements in Intersection) We are looking for pairs of subsets $(A, B)$ such that $n(A \cap B) = 2$. We can achieve this by following two sub-steps:

Sub-step 3.1: Choose the elements that will be in the intersection ($A \cap B$). We need to select exactly 2 elements from the 5 available elements in set $S$ to form the intersection $A \cap B$. The number of ways to do this is given by the combination formula $^n C_k$: $\text{Number of ways to choose elements for } A \cap B = ^5 C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$ Let's assume these two chosen elements are now fixed as members of $A \cap B$. This means for these 2 elements, their "fate" is decided (they are in $A$ and in $B$).

Sub-step 3.2: Distribute the remaining elements. After selecting 2 elements for the intersection, there are $5 - 2 = 3$ elements remaining in the set $S$. For each of these 3 remaining elements, it cannot be in the intersection $A \cap B$ (because we have already fixed the elements that are in the intersection). Therefore, each of these 3 remaining elements must belong to one of the other three possible regions:

• In $A$ but not in $B$ ($A \setminus B$)
• In $B$ but not in $A$ ($B \setminus A$)
• Neither in $A$ nor in $B$ ($S \setminus (A \cup B)$) So, for each of the 3 remaining elements, there are 3 independent choices. The total number of ways to distribute these 3 remaining elements is: $\text{Number of ways to distribute remaining elements} = 3 \times 3 \times 3 = 3^3 = 27$

Combining Sub-steps for Favorable Outcomes: To get the total number of favorable outcomes, we multiply the number of ways to choose the intersection elements by the number of ways to distribute the remaining elements: $\text{Number of Favorable Outcomes} = (^5 C_2) \times 3^3 = 10 \times 27 = 270$

Step 4: Calculate the Required Probability Now, we use the probability formula: $P(A \cap B \text{ has 2 elements}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$ $P(A \cap B \text{ has 2 elements}) = \frac{270}{2^{10}}$ To simplify the fraction, we can divide the numerator and denominator by 2: $P(A \cap B \text{ has 2 elements}) = \frac{135}{2^9}$

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Common Mistakes & Tips

• Miscounting Total Outcomes: A common mistake is to count $2^n \times 2^n$ as the total number of outcomes, which is correct, but sometimes students might think of $^N C_2$ where $N=2^n$ if they are considering choosing two distinct subsets from the $2^n$ possible subsets. The phrasing "two randomly selected subsets" typically implies ordered pairs $(A, B)$ where $A$ and $B$ are chosen independently. The "fate of each element" method ($4^n$) inherently counts ordered pairs.
• Incorrect Distribution of Remaining Elements: Ensure that once elements are assigned to a specific region (like the intersection), they cannot be assigned to other regions. The remaining elements must be distributed among the remaining allowed regions.
• Overlooking Independence: Remember that the choices for each element's placement are independent of the choices for other elements. This allows for multiplication of possibilities.

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Summary

This problem required us to calculate the probability that two randomly selected subsets of a 5-element set have exactly two elements in their intersection. We first determined the total number of ways to form two ordered subsets by considering the 4 possible regions each of the 5 elements could belong to, resulting in $4^5 = 2^{10}$ total outcomes. Next, we calculated the number of favorable outcomes by first choosing 2 elements for the intersection ($^5 C_2$ ways) and then distributing the remaining 3 elements into the 3 non-intersection regions ($3^3$ ways). This yielded $10 \times 27 = 270$ favorable outcomes. Finally, dividing the favorable outcomes by the total outcomes gave the probability $\frac{270}{2^{10}} = \frac{135}{2^9}$.

The final answer is $\boxed{\frac{135}{2^9}}$ which corresponds to option (A).