Let S be a set of 5 elements and P(S) denote the power set of S. Let E be an event of choosing an ordered pair (A, B) from the set P(S) x P(S) such that $$A\cap B=\phi.$$ If
the probability of the event E is $$\frac{3^{p}}{2^{q}}$$, where p,q $$\in$$ N, then p + q is equal to __________

$$S$$ has 5 elements. We choose an ordered pair $$(A, B)$$ from $$P(S) \times P(S)$$. We need the probability that $$A \cap B = \phi$$.

• Total Number of Pairs: Each set $$A$$ and $$B$$ is a subset of $$S$$. Since $$S$$ has 5 elements, $$P(S)$$ has $$2^5$$ elements. The total number of ordered pairs $$(A, B)$$ is $$2^5 \times 2^5 = 2^{10}$$.

• Favorable Outcomes: For each element $$x \in S$$, there are 4 possibilities for its "membership" in the pair $$(A, B)$$:

1. $$x \in A$$ and $$x \in B$$

2. $$x \in A$$ and $$x \notin B$$

3. $$x \notin A$$ and $$x \in B$$

4. $$x \notin A$$ and $$x \notin B$$

• For $$A \cap B = \phi$$, the first case ($$x$$ in both) is impossible. This leaves 3 choices for each of the 5 elements.

• Total favorable pairs $$= 3^5$$.

• Probability: $$P(E) = \frac{3^5}{2^{10}}$$.

• Comparing this to $$\frac{3^p}{2^q}$$, we get $$p = 5$$ and $$q = 10$$.

• Final Answer: $$p + q = 5 + 10 = \mathbf{15}$$.