Question:
Let $S$ be a set of $5$ elements and $P(S)$ denote the power set of $S$.
Let $E$ be the event of choosing an ordered pair $(A,B)$ from $P(S)\times P(S)$ 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
Let $S$ be a set of $5$ elements and $P(S)$ denote the power set of $S$.
Let $E$ be the event of choosing an ordered pair $(A,B)$ from $P(S)\times P(S)$ 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
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When choosing disjoint subsets, think in terms of independent choices for each element.
Updated On: Feb 24, 2026
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Correct Answer: 15
Solution and Explanation
Step 1: Find total outcomes.
\[ |P(S)|=2^5=32 \] \[ \text{Total ordered pairs}=32\times32=2^{10} \] Step 2: Count favourable outcomes.
Each element of $S$ can be: \[ \text{in }A,\ \text{in }B,\ \text{or in neither} \] So, \[ \text{Favourable outcomes}=3^5 \] Step 3: Compute probability.
\[ P(E)=\frac{3^5}{2^{10}} \] Thus, \[ p=5,\quad q=10 \] Step 4: Final answer.
\[ p+q=15 \]
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