Animated Solution for Mathematics - Probability: If $C$ and $D$ are two events such that $C\subset D$ and $P(D)\ne 0$, then the correct statement among the following is

Imagine you are standing in a vast, open field. In the world of probability, we call this field our sample space, denoted by $S$. It is the grand stage where every possible outcome of an experiment resides.

Within this space, we have two events, $C$ and $D$. The problem provides a crucial constraint: $C\subset D$.

This is a geometric reality. It means that every single outcome that makes event $C$ happen is also an outcome that makes event $D$ happen. If $C$ occurs, $D$ is guaranteed to occur.

We are asked to evaluate $P(C|D)$, the probability of $C$ given that $D$ has already occurred. This is the heart of conditional probability, defined by the following formula:

This formula tells us that when we know $D$ has occurred, our "universe" has shrunk. We no longer care about the parts of $S$ that are outside $D$; our new sample space is restricted to $D$.

Now, consider the numerator: $P(C\cap D)$. Because we know $C\subset D$, the intersection of $C$ and $D$ is simply $C$ itself.

Every point in $C$ is already contained within $D$. Therefore, the overlap is just $C$, which simplifies our expression:

Substituting this back into our conditional probability formula, we obtain:

We are given that $P(D) eq 0$. Since $P(D)$ is a probability, we know that $0<P(D)\le 1$.

Consider the term $\frac{1}{P(D)}$. In all cases where $P(D)\le 1$, the reciprocal must satisfy the following inequality:

We have established that $P(C|D)=P(C)\cdot \frac{1}{P(D)}$. Since $\frac{1}{P(D)}\ge 1$, multiplying $P(C)$ by this factor can only make it larger or keep it the same.

Therefore, we arrive at the final result:

This result is intuitive. If you know that a larger event $D$ has occurred, and $C$ is a subset of $D$, the likelihood of $C$ happening is higher than it was before you knew anything about $D$. You have eliminated the "noise" outside of $D$, making the occurrence of $C$ more probable.