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NVAEasyJEE 2026Probability Basics

JEE Mathematics 2026 Question with Solution

Let SS be a set of 55 elements and P(S)P(S) denote the power set of SS. Let EE be the event of choosing an ordered pair (A,B)(A,B) from P(S)×P(S)P(S)\times P(S) such that A∩B=∅A\cap B=\varnothing. If the probability of the event EE is 3p2q\dfrac{3^p}{2^q}, where p,q∈Np,q\in\mathbb{N}, then p+qp+q is equal to

Answer

Correct answer:15

Step-by-step solution

Standard Method

Given: SS is a set with 55 elements. We choose an ordered pair (A,B)(A,B) from P(S)×P(S)P(S)\times P(S) such that A∩B=∅A\cap B=\varnothing.

Find: p+qp+q if

P(E)=3p2qP(E)=\frac{3^p}{2^q}

Step 1: Find total outcomes. Since ∣P(S)∣=25=32|P(S)|=2^5=32, the total number of ordered pairs is

32×32=21032\times 32 = 2^{10}

Step 2: Count favourable outcomes. For each element of SS, it can be:

  • in AA,
  • in BB, or
  • in neither,

because A∩B=∅A\cap B=\varnothing. Hence each element has 33 valid choices, so the number of favourable ordered pairs is

353^5

Step 3: Compute the probability.

P(E)=35210P(E)=\frac{3^5}{2^{10}}

Comparing with

3p2q\frac{3^p}{2^q}

we get

p=5,q=10p=5, \qquad q=10

Conclude: Therefore,

p+q=15p+q=15

So the required answer is 1515.

Common mistakes

  • Assuming each element has 44 choices: in AA only, in BB only, in both, or in neither. This is wrong because the condition A∩B=∅A\cap B=\varnothing forbids an element from being in both sets. Use only 33 choices per element.

  • Counting unordered pairs instead of ordered pairs. This is wrong because (A,B)(A,B) and (B,A)(B,A) are different outcomes in P(S)×P(S)P(S)\times P(S). The sample space must use ordered pairs.

  • Using 252^5 as the number of favourable outcomes. This is wrong because the condition involves both subsets together, not one subset alone. Count choices element-wise across both AA and BB to get 353^5.

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