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) × P(S) such that A ∩ B = ∅. If the probability of the event E is 3^p / 2^q, where p, q ∈ N, then p + q is equal to

Let S be a set of 5 elements and P(S) denote the power set of S

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

Correct Answer: 15

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.

Explanation

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