Animated Solution for Mathematics - Probability: Let $A$ and $B$ be two non-null events such that $A\subset B$. Then, which of the following statements is always correct ?

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the elegant, logical architecture of probability.

We are given two events, $A$ and $B$, with the condition that $A\subset B$. This is our starting point, our map.

Imagine a vast, rectangular universe—our sample space. Inside this universe, there is a blue circle representing event $B$. Because $A\subset B$, the event $A$ is a smaller circle, nestled entirely within the blue circle of $B$.

This is the geometric reality of our problem. Every time $A$ happens, $B$ is already there, waiting. This visual is the key to everything that follows.

Now, we turn our attention to the concept of conditional probability, $P(A|B)$. This is the probability of $A$ occurring, given that we already know $B$ has occurred.

The formal definition is our guiding light:

This formula is not just a collection of symbols; it is a ratio of possibilities. It tells us to look only at the outcomes where $B$ happens, and then see what fraction of those outcomes also include $A$.

But wait! Look back at our Venn diagram. Since $A$ is completely inside $B$, the intersection $A\cap B$ is simply $A$ itself. There is no part of $A$ that exists outside of $B$.

Thus, $P(A\cap B)=P(A)$. Our formula simplifies beautifully to:

We have stripped away the complexity and arrived at the core relationship.

Now, we must be careful. We are dealing with inequalities, and this is where many students stumble. We know that for any event $B$, its probability $P(B)$ is bounded: $0<P(B)\le 1$.

We are given that $B$ is non-null, so $P(B)>0$. Now, consider the reciprocal. When we take the reciprocal of an inequality involving positive numbers, the direction of the inequality flips.

So, $\frac{1}{P(B)}\ge 1$. This is the pivot point of our argument. We are essentially saying that dividing by a probability (which is at most 1) is the same as multiplying by a number greater than or equal to 1.

Finally, we multiply both sides of this inequality by $P(A)$. Since $P(A)$ is also positive, the inequality sign remains steady. We get:

And what is the left side of this expression? It is exactly our conditional probability $P(A|B)$! Thus, we arrive at the profound conclusion: $P(A|B)\ge P(A)$.

Why does this make sense? Think about it. If you know that $B$ has occurred, you have restricted your sample space. You are no longer looking at the entire universe; you are looking only at the world of $B$.

Since $A$ is a part of $B$, and you have removed all the outcomes outside of $B$ that were not part of $A$, the relative likelihood of $A$ occurring must increase or stay the same. It cannot decrease.

This is the beauty of probability—it aligns perfectly with our intuition once we strip away the abstraction. You have successfully navigated the logic of subsets and conditional probability. Keep this clarity with you as you tackle the next challenge.