Animated Solution for Mathematics - Probability: $A$ is a set containing $n$ elements. A subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$. A subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.

Mastering Set Theory

Welcome, future engineer. Today, we are not just solving a probability problem; we are learning a mindset. In JEE Advanced, the most daunting problems are often those that look like they require massive, complex summations.

But the secret to mastering these is to stop looking at the 'whole' and start looking at the 'individual.' Let us dive into this problem of sets $P$ and $Q$ within a universal set $A$.

When you first read the problem, your brain might try to calculate the total number of subsets of $A$, which is $2^n$, and then try to find the number of pairs $(P,Q)$ such that $P\cap Q=\varnothing$. While mathematically sound, this path is a labyrinth.

You would be dealing with combinations, summations, and binomial coefficients that could easily lead to a calculation error. Instead, let us adopt the 'Element-wise Perspective.'

Imagine you are standing before the set $A=\{a_1,a_2,\dots ,a_n\}$. Instead of worrying about the entire set, let us focus on one single element, $a_i$. This element is our protagonist.

When we choose subset $P$ and then subset $Q$, what can happen to $a_i$? For any element $a_i$, there are exactly four mutually exclusive possibilities:

1. $a_i\in P$ and $a_i\in Q$ (The element is in both) 2. $a_i\in P$ and $a_i otin Q$ (The element is only in $P$) 3. $a_i otin P$ and $a_i\in Q$ (The element is only in $Q$) 4. $a_i otin P$ and $a_i otin Q$ (The element is in neither)

Since each element has these 4 independent choices, and there are $n$ elements in total, the total number of ways to form the subsets $P$ and $Q$ is $4\times 4\times \cdots \times 4$ ($n$ times). This gives us $4^n$, which is our sample space.

Now, let us look at the constraint: $P\cap Q=\varnothing$. This is the 'forbidden' condition. It tells us that no element is allowed to be in both $P$ and $Q$ simultaneously.

Looking back at our four possibilities, the first state ($a_i\in P$ and $a_i\in Q$) is strictly prohibited. If we allow this state, we violate the condition of having no common elements. So, for each element $a_i$, we are left with exactly 3 valid states:

1. $a_i\in P$ and $a_i otin Q$ 2. $a_i otin P$ and $a_i\in Q$ 3. $a_i otin P$ and $a_i otin Q$

Since each of the $n$ elements has exactly 3 valid choices, the total number of favorable ways to choose $P$ and $Q$ such that they are disjoint is $3\times 3\times \cdots \times 3$ ($n$ times), which is $3^n$.

Finally, the probability is simply the ratio of favorable ways to total ways:

This simplifies beautifully to:

Look at the elegance of the result. We didn't need to perform complex summations or worry about the size of $n$. By focusing on the fate of a single element, we reduced a combinatorial nightmare into a simple geometric probability.

This is the essence of JEE Advanced physics and mathematics: finding the simplest, most fundamental perspective. Keep this 'element-wise' strategy in your toolkit. Whether it is probability, combinatorics, or even electrostatics, always ask yourself: 'What is happening to a single unit?' You will be surprised at how often the answer becomes clear.