Animated Solution for Mathematics - Probability: If $A$ and $B$ are two events such that $P(A)>0$, and $P(B)\ne 1$, then $P\left(\frac{\bar{A}}{\bar{B}}\right)$ is equal to

We begin with the fundamental definition of conditional probability. When we write $P(\bar{A}|\bar{B})$, we are asking a specific question: "Given that we are restricted to the world where $B$ does not happen, what is the probability that $A$ also does not happen?"

Mathematically, this is defined as:

This equation is our anchor. The denominator, $P(\bar{B})$, represents our new "universe." Because we are given $P(B) eq 1$, we know $P(\bar{B})=1-P(B)>0$, ensuring we are safe from the abyss of division by zero.

Now, look at the numerator: $P(\bar{A}\cap \bar{B})$. This is where many students stumble, as they often try to calculate probabilities of individual events instead of utilizing set theory.

We invoke De Morgan's Law, a beautiful symmetry in logic. It tells us that the intersection of the complements of two sets is identical to the complement of their union:

Imagine the Venn diagram. The region $\bar{A}$ is everything outside circle $A$, and the region $\bar{B}$ is everything outside circle $B$. The intersection is the region that is outside both circles, which is precisely the entire sample space $S$ minus the combined area of $A$ and $B$.

Thus, we can write:

We are almost at the finish line. We know that for any event $E$, the probability of its complement is $P(\bar{E})=1-P(E)$.

Applying this to our union event, we get:

Now, we substitute this back into our original conditional probability expression:

Look at this result. It is elegant, concise, and logically sound. It perfectly matches the third option provided in the problem.

By visualizing the sample space and applying set theory, we transformed a complex conditional probability into a simple, manageable expression. Keep practicing this level of rigor, and you will find that even the most intimidating problems begin to unravel before your eyes.