Aspire Faculty ID #18084 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
JEE Main 2026 (24 January Evening Shift)

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 = \emptyset $. If the probability of the event $ E $ is $ \frac{3^p}{2^q} $, where $ p, q \in \mathbb{N} $, then $ p + q $ is equal to

Solution

$ S = {a, b, c, d, e} $ $ P(S) $ contains 32 elements both set $ A $ and set $ B $ are subsets of $ P(S) $ Every element has 4 choices Favourable cases $ = 3^5 $ Total cases $ = 4^5 $ $ P = \frac{3^5}{4^5} = \frac{3^5}{2^{10}} $ $ p = 5,; q = 10 $ $ p + q = 15 $
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