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

Hint:

When choosing disjoint subsets, think in terms of independent choices for each element.

Answer 1

Correct Answer: 15

To solve this problem, we aim to determine the probability of selecting an ordered pair $(A,B)$ from the power set of a finite set $S$ with 5 elements such that $A\cap B=\varnothing$. The power set $P(S)$ of a set with $n$ elements has $2^n$ subsets. Thus, for $S$ with 5 elements, $|P(S)|=2^5=32$.
Therefore, the Cartesian product $P(S)\times P(S)$ contains $32\times 32=1024$ ordered pairs. Now consider the event $E$ requiring $A\cap B=\varnothing$. Each element of $S$ can independently belong to $A$, $B$, or neither, but it cannot belong to both since $A\cap B$ must be empty. For each element in $S$, there are thus 3 choices, leading to $3^5=243$ such pairs $(A,B)$.
Consequently, the probability of $E$ is given by: \[\frac{243}{1024}=\frac{3^5}{2^{10}}.\]
Here, $p=5$ and $q=10$, so $p+q=5+10=15$.
This result, $p+q=15$