Q. 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
\(P(S) \times P(S)\) such that \(A \cap B = \varnothing\).
If the probability of the event \(E\) is
\(\dfrac{3^p}{2^q}\), where \(p, q \in \mathbb{N}\),
then \(p + q\) is equal to
Explanation
The total number of elements in set \(S\) is 5.
Number of subsets of \(S\):
\[
|P(S)| = 2^5 = 32
\]
Total number of ordered pairs \((A,B)\) from \(P(S) \times P(S)\):
\[
32 \times 32 = 1024
\]
Now count favorable cases where \(A \cap B = \varnothing\).
For each element of \(S\), it can belong to:
- Only in \(A\)
- Only in \(B\)
- In neither \(A\) nor \(B\)
So each element has 3 independent choices.
\[
\text{Favorable outcomes} = 3^5
\]
Hence, probability of event \(E\) is
\[
P(E) = \frac{3^5}{2^{10}}
\]
Comparing with \(\dfrac{3^p}{2^q}\),
\[
p = 5,\quad q = 10
\]
\[
p + q = 5 + 10 = \boxed{15}
\]
Therefore, the correct answer is 15.