Animated Solution for Mathematics - Probability: If two events $A$ and $B$ are such that $P(A^c)=0.3,P(B)=0.4$ and $P(A\cap B^c)=0.5$, then $P(B|A\cup B^c)=.........$

A Journey into Probability

Welcome, future engineer. Today, we are not just solving a probability problem; we are mapping the landscape of uncertainty.

Probability is often taught as a series of dry formulas, but at the JEE Advanced level, it is a visual language. When you look at a problem involving events $A$ and $B$, I want you to stop seeing variables and start seeing regions on a map.

Let us embark on this journey together.

The Map of the Universe

Imagine a rectangular box representing the universal set $U$. Inside, we have two circles, $A$ and $B$.

The beauty of this problem lies in the information given: $P(A^c)=0.3$, $P(B)=0.4$, and $P(A\cap B^c)=0.5$.

Before we touch a single equation, let us find our bearings. We are given $P(A^c)=0.3$. The complement rule is our first tool: $P(A)=1-P(A^c)$.

Thus, $P(A)=1-0.3=0.7$. We now know the total weight of circle $A$.

Next, look at $P(A\cap B^c)=0.5$. This is the 'A-only' region—the part of $A$ that does not touch $B$. Geometrically, this is the crescent shape of $A$ excluding the overlap.

If we know the total $P(A)$ is $0.7$ and the 'A-only' part is $0.5$, the overlap $P(A\cap B)$ must be the difference:

We have successfully unlocked the intersection.

Constructing the Denominator

Now, we turn our attention to the conditional probability we need to find: $P(B|A\cup B^c)$. By the definition of conditional probability, this is:

Let us tackle the denominator first: $P(A\cup B^c)$. We use the addition theorem for probability:

We already have $P(A)=0.7$ and $P(A\cap B^c)=0.5$. We just need $P(B^c)$. Since $P(B)=0.4$, then $P(B^c)=1-0.4=0.6$.

Substituting these values, we get $P(A\cup B^c)=0.7+0.6-0.5=0.8$. Our denominator is ready.

The Elegance of the Numerator

Now, the numerator: $P(B\cap (A\cup B^c))$. This looks intimidating, but let us apply the distributive law of sets.

Just as we distribute multiplication over addition in algebra, we distribute the intersection over the union:

Here is the moment of clarity. What is $B\cap B^c$? It is the set of outcomes that are in $B$ AND not in $B$. That is impossible; it is the empty set $\varnothing$.

Therefore, the expression collapses to:

We already calculated $P(A\cap B)=0.2$ in our first phase. The complexity vanishes, leaving us with a simple ratio.

The Final Calculation

We have arrived at the finish line. We have our numerator, $0.2$, and our denominator, $0.8$.

The conditional probability is:

Look at that result. It is clean, precise, and elegant. You did not just calculate a number; you navigated through set theory, applied the distributive law, and visualized the geometry of probability.

This is the essence of JEE Advanced physics and mathematics. Keep this mindset—visualize first, calculate second—and you will conquer any problem that comes your way. The final answer is $1/4$.