Animated Solution for Mathematics - Probability: $P(A\cup B)=P(A\cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is .........

Unlocking the Mystery of $P(A\cup B)=P(A\cap B)$

Welcome, future engineers. Today, we are not just solving an equation; we are peeling back the layers of a fundamental truth in probability theory.

Often, when we look at a problem like $P(A\cup B)=P(A\cap B)$, our instinct is to dive straight into the algebra. But I want you to pause. I want you to visualize.

Before we touch a single variable, let us stand on the firm ground of geometry.

The Venn Diagram Perspective

Imagine you are looking at a Venn diagram. You have two circles, $A$ and $B$.

The union, $A\cup B$, is the entire landscape covered by both circles. It is the sum total of everything that happens in either $A$ or $B$.

Now, look at the intersection, $A\cap B$. That is the small, overlapping heart of the diagram.

The problem presents us with a paradox: the entire landscape is equal to the heart. How can the whole be equal to just a small part?

This is the spark of our investigation. It tells us that the regions outside the intersection—the 'A only' and 'B only' zones—must be effectively empty. They must have no probability mass.

The Addition Theorem

To translate this geometric intuition into the language of mathematics, we reach for our most reliable tool: the Addition Theorem of Probability. It is the bridge between the visual and the analytical.

We know that:

This formula is elegant because it corrects for the double-counting of the intersection. When we add the probability of $A$ and the probability of $B$, we count the overlap twice.

Subtracting it once restores the balance. This is the bedrock upon which we will build our proof.

The Algebraic Rearrangement

Now, let us apply our given condition. We are told that $P(A\cup B)=P(A\cap B)$.

Let us substitute this into our Addition Theorem. The equation transforms into:

This looks simple, but it is the turning point. We want to isolate the relationship between $P(A)$ and $P(B)$.

By adding $P(A\cap B)$ to both sides, we get:

Or, if we prefer to see it as a sum equal to zero, we write:

Do not rush this step. Ensure your signs are correct. A single misplaced negative sign here would derail the entire logic.

The Logic of Non-Negativity

Here is where the magic happens. We have the expression $P(A)+P(B)-2P(A\cap B)=0$.

Let us split that $-2P(A\cap B)$ into two separate terms: $-P(A\cap B)$ and $-P(A\cap B)$. Now, group them strategically:

Look closely at these brackets. $P(A)-P(A\cap B)$ is the probability of $A$ occurring without $B$, which we write as $P(A\setminus B)$. Similarly, $P(B)-P(A\cap B)$ is $P(B\setminus A)$.

Our equation is now:

Why is this profound? Because probability is non-negative. $P(E)\ge 0$ for any event $E$.

We have the sum of two non-negative numbers equaling zero. In the realm of real numbers, the only way for the sum of two non-negative values to be zero is if each value is zero itself.

Therefore, $P(A\setminus B)=0$ and $P(B\setminus A)=0$. The 'only A' and 'only B' regions are empty!

The Final Equality

Since $P(A\setminus B)=0$, it follows that $P(A)=P(A\cap B)$. Similarly, since $P(B\setminus A)=0$, it follows that $P(B)=P(A\cap B)$.

If both $P(A)$ and $P(B)$ are equal to the same intersection, they must be equal to each other. Thus:

We have arrived at our destination. We started with a geometric paradox, used the Addition Theorem to bridge the gap, and employed the non-negativity of probability to seal the proof.

This is the beauty of JEE mathematics—it is not just about calculation; it is about logical deduction. Keep this clarity with you as you tackle more complex problems.