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

Correct answer: 0 S has 5 elements, so P(S) has 2^5 = 32 subsets. Total ordered pairs from P(S) × P(S) is 32^2 = 1024 For pairs (A, B) with A B = : each of the 5 elements has 3 choices - in A only, in B only, or in neither Number of favorable outcomes = 3^5 = 243 Probability = (243)/(1024) = 3^5 2^ 10 With p = 5 and q = 10 , we get p + q = 15

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Probability

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

Let S be a set of 5 elements and

\mathrm{P}(\mathrm{S})

denote the power set of S. Let E be an event of choosing an ordered pair (

\mathrm{A}, \mathrm{B}

) from the set

\mathrm{P}(\mathrm{S}) \times \mathrm{P}(\mathrm{S})

such that

\mathrm{A} \cap \mathrm{B}=\emptyset

. If the probability of the event

E

\frac{3^{p}}{2^{q}}

, where

p, q \in N

, then

p+q

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

S

has 5 elements, so

P(S)

has

2^5 = 32

subsets.
Total ordered pairs from

P(S) \times P(S)

32^2 = 1024

For pairs

(A, B)

with

A \cap B = \emptyset

: each of the 5 elements has 3 choices - in

A

only, in

B

only, or in neither
Number of favorable outcomes =

3^5 = 243

Probability

= \frac{243}{1024} = \frac{3^5}{2^{10}}

With

p = 5

and

q = 10

, we get

p + q = 15

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