Animated Solution for Mathematics - Probability: If $P(B)=\frac{3}{4},P(A\cap B\cap \bar{C})=\frac{1}{3}$ and $P(\bar{A}\cap B\cap \bar{C})=\frac{1}{3}$, then $P(B\cap C)$ is

A Journey into Set $B$

Welcome, fellow explorer of the mathematical universe! Today, we aren't just solving a probability problem; we are conducting a survey of a landscape.

Imagine you are standing before a Venn diagram, a map of three overlapping territories: $A$, $B$, and $C$. Our mission is to understand the composition of territory $B$. We know its total size is $P(B)=\frac{3}{4}$, but we need to find the specific value of the intersection $P(B\cap C)$.

The Art of Partitioning

In JEE problems, the secret to success is often 'divide and conquer.' We want to break down the large, intimidating set $B$ into smaller, manageable pieces.

Think of set $B$ as a circular plot of land. We can slice this land using the boundary of set $C$. This gives us two distinct, non-overlapping regions: the part of $B$ that is inside $C$ (which is $B\cap C$) and the part of $B$ that is outside $C$ (which is $B\cap \bar{C}$).

But wait, the region $B\cap \bar{C}$ is still quite large. Let's refine our survey. We can use set $A$ to slice this region further.

By looking at the intersection of $A$ and $B$ outside $C$, we identify the region $A\cap B\cap \bar{C}$. What remains is the part of $B$ that is outside both $A$ and $C$, which we denote as $\bar{A}\cap B\cap \bar{C}$.

The Algebra of Logic

Now, we have successfully partitioned set $B$ into three disjoint, non-overlapping puzzle pieces: 1. $B\cap C$ (Our target) 2. $A\cap B\cap \bar{C}$ (The red region) 3. $\bar{A}\cap B\cap \bar{C}$ (The blue region)

Because these regions are disjoint, the Addition Theorem of probability tells us that the total probability of $B$ is simply the sum of these parts:

This equation is our master key. It connects the knowns—the total probability of $B$ and the sizes of the two outer regions—to our unknown target, $P(B\cap C)$.

The Final Calculation

Let's plug in the values provided in the problem statement:

Substituting these into our master equation, we get:

Simplifying the right side, we have:

Now, we isolate our target variable by subtracting $\frac{2}{3}$ from both sides:

To perform this subtraction, we find a common denominator, which is $12$:

The Elegance of Visualization

And there we have it! The probability of $B\cap C$ is $\frac{1}{12}$.

By visualizing the sets and partitioning them into disjoint regions, we transformed a potentially confusing problem into a simple, elegant arithmetic exercise. This is the power of the JEE mindset—don't just calculate; visualize, partition, and conquer.

Keep this technique in your toolkit, and you'll find that even the most complex probability problems start to reveal their secrets.