Power Set Size Calculation

Given a set $S$ with 5 elements, $|S| = 5$. The power set $\text{P}(S)$ contains all possible subsets of $S$. The number of elements in the power set is given by $2^{|S|}$.

• Number of elements in $\text{P}(S) = 2^5 = 32$.

Total Ordered Pairs Calculation

The problem involves choosing an ordered pair $(A, B)$ from the Cartesian product $\text{P}(S) \times \text{P}(S)$. The total number of such ordered pairs is the product of the sizes of the individual sets.

• Total ordered pairs $= |\text{P}(S)| \times |\text{P}(S)| = 32 \times 32 = 1024$.
• This can also be expressed as $2^5 \times 2^5 = 2^{10}$.

Disjoint Pairs Calculation

The event $\text{E}$ requires that the chosen subsets $A$ and $B$ are disjoint, meaning $A \cap B = \emptyset$. For each element $x \in S$, there are three possibilities regarding its membership in sets $A$ and $B$ such that they remain disjoint:

• $x$ is in $A$ only ($x \in A, x \notin B$).
• $x$ is in $B$ only ($x \notin A, x \in B$).
• $x$ is in neither $A$ nor $B$ ($x \notin A, x \notin B$).

Since there are 5 elements in $S$, and each element has 3 independent choices for its placement relative to $A$ and $B$ to ensure disjointness, the total number of pairs $(A, B)$ satisfying $A \cap B = \emptyset$ is $3^{|S|}$.

• Number of disjoint pairs $(A, B) = 3^5 = 243$.

Probability Calculation

The probability of event $\text{E}$ is the ratio of the number of favorable outcomes (disjoint pairs) to the total number of possible outcomes (all ordered pairs).

• $P(\text{E}) = \frac{\text{Number of disjoint pairs}}{\text{Total ordered pairs}} = \frac{3^5}{2^{10}}$.

Determining p and q

The problem states that the probability of event $\text{E}$ is given in the form $\frac{3^p}{2^q}$, where $p, q \in \mathbb{N}$.

• Comparing $P(\text{E}) = \frac{3^5}{2^{10}}$ with $\frac{3^p}{2^q}$, we find that $p = 5$ and $q = 10$.

Final Sum Calculation

We need to find the value of $p + q$.

• $p + q = 5 + 10 = 15$.