Animated Solution for Mathematics - Probability: $A,B,C$ are events such that $P(A)=0.3,P(B)=0.4,P(C)=0.8,P(AB)=0.08,P(AC)=0.28,P(ABC)=0.09$. If $P(A\cup B\cup C)\ge 0.75$, then show that $P(BC)$ lies in the interval $0.23\le x\le 0.48$.

Unlocking the Inclusion-Exclusion Principle

Welcome, fellow traveler on the path of JEE Advanced mathematics. Today, we are not just solving a problem; we are exploring the elegant, interconnected nature of probability.

Often, when we look at events $A$, $B$, and $C$, we see them as separate entities. But in the world of probability, they are dancers on a stage, constantly overlapping, influencing, and constraining one another.

Our goal today is to find the 'zone of possibility' for the intersection of $B$ and $C$, given the constraints of their union.

The Geometry of Overlap

Imagine you have three circles on a piece of paper. If you want to calculate the total area covered by these circles—the union $P(A\cup B\cup C)$—you cannot simply add their individual areas.

If you add $P(A)+P(B)+P(C)$, you have counted the pairwise intersections like $P(AB)$ twice. To fix this, we subtract them.

But wait! By subtracting the pairwise intersections, we have now removed the central region—the intersection of all three, $P(ABC)$—one too many times. So, we must add it back.

This is the heart of the Inclusion-Exclusion Principle:

This formula is a correction mechanism for the human tendency to over-count. Let us breathe life into this equation with the values provided: $P(A)=0.3$, $P(B)=0.4$, $P(C)=0.8$, $P(AB)=0.08$, $P(AC)=0.28$, and $P(ABC)=0.09$.

The Algebraic Collapse

Now, let us substitute these values into our master equation. We are simplifying the chaos into order:

Let us group the constants. First, the positive terms: $0.3+0.4+0.8+0.09=1.59$. Now, the negative terms: $0.08+0.28=0.36$.

When we subtract these, we get $1.59-0.36=1.23$. Suddenly, our complex expression collapses into something beautifully simple:

This is the moment of clarity. We have successfully linked the union of our three events directly to the unknown intersection $P(BC)$.

The Constraint of Reality

Here is where the physics of probability takes over. We are told that $P(A\cup B\cup C)\ge 0.75$. But we also know an implicit truth: the probability of any event cannot exceed $1$.

Therefore, we have a 'sandwich' constraint:

Substitute our simplified expression into this inequality:

We are now solving for $P(BC)$. To isolate it, we subtract $1.23$ from all parts of the inequality:

The Final Transformation

We are almost at the finish line. We have $-P(BC)$ bounded between $-0.48$ and $-0.23$. To find $P(BC)$, we multiply the entire inequality by $-1$.

Remember, when you multiply an inequality by a negative number, the direction of the inequality signs must flip. It is a fundamental rule of algebra that protects the integrity of the number line:

Rearranging this to the standard format, we arrive at our destination:

We have proven it. The probability of $B$ and $C$ occurring together is not a fixed point, but a range of possibilities dictated by the constraints of the union. You have navigated the Inclusion-Exclusion Principle, handled the algebraic constraints, and emerged with the correct interval.