Animated Solution for Mathematics - Probability: It is given that the events $A$ and $B$ are such that $P(A)=1/4,P(A|B)=1/2$ and $P(B|A)=2/3$. Then $P(B)$ is

Welcome, future engineers! Today, we are not just solving a probability problem; we are exploring the architecture of uncertainty. When you look at a problem involving $P(A)$, $P(A|B)$, and $P(B|A)$, it is easy to feel overwhelmed by the notation.

Imagine the sample space $S$ as a vast, open field. Within this field, we have two events, $A$ and $B$, represented by two overlapping circles. The area where they overlap is the intersection, $P(A\cap B)$.

This intersection is the most important piece of the puzzle—it is the bridge that connects the world of $A$ to the world of $B$.

We are given $P(A)=\frac{1}{4}$ and $P(B|A)=\frac{2}{3}$. The definition of conditional probability is our compass here:

This formula tells us that the probability of $B$ happening, given that $A$ has already occurred, is simply the ratio of the shared area to the area of $A$ itself. If we rearrange this, we get the beautiful, simple equation:

By multiplying these two values, we are effectively calculating the 'size' of the overlap. Substituting our values, we get:

We have successfully crossed the bridge!

Now that we have the intersection, the rest of the problem unfolds like a well-choreographed dance. We are given $P(A|B)=\frac{1}{2}$. Using the definition of conditional probability again, we know that:

We want to find $P(B)$, so let's isolate it:

We already know $P(A\cap B)=\frac{1}{6}$ and $P(A|B)=\frac{1}{2}$. Substituting these in, we get:

To divide by a fraction, we multiply by its reciprocal:

Simplifying this, we arrive at our final answer:

It is elegant, it is logical, and it is correct. Remember, in JEE, when you see conditional probabilities, do not panic. Find the intersection, build your bridge, and the path to the solution will always reveal itself.