Animated Solution for Mathematics - Probability: $A$ and $B$ are two independent events. $C$ is event in which exactly one of $A$ or $B$ occurs. Prove that $P(C)\ge P(A\cup B)P(\bar{A}\cap \bar{B})$.

Welcome, future engineers. Today, we are peeling back the layers of probability to see how events interact in the abstract space of outcomes. We are given two independent events, $A$ and $B$.

Independence is a powerful concept; it tells us that the occurrence of $A$ gives us zero information about the occurrence of $B$. Mathematically, this is our golden key:

Let us define our variables to make the battlefield clear. Let $P(A)=x$ and $P(B)=y$. Consequently, the intersection is $xy$.

Now, consider event $C$, where exactly one of $A$ or $B$ occurs. Imagine a Venn diagram; we want the parts of the circles that do not overlap.

This is the symmetric difference. If we take the entire union $P(A\cup B)$ and subtract the intersection $P(A\cap B)$, we are left with exactly the regions we want:

Using our variables, we know $P(A\cup B)=x+y-xy$. Thus, the expression for $P(C)$ becomes:

Now, look at the right-hand side (RHS) of our inequality: $P(A\cup B)P(\bar{A}\cap \bar{B})$. That second term, $P(\bar{A}\cap \bar{B})$, can be simplified using De Morgan's Law.

De Morgan's Law tells us that $\bar{A}\cap \bar{B}=\overline{A\cup B}$. Since the probability of the complement is $1$ minus the probability of the event, we have:

Let us simplify our notation by setting $z=P(A\cup B)$. Now, the right-hand side becomes:

A Strategy for Victory

We need to prove $P(C)\ge P(A\cup B)P(\bar{A}\cap \bar{B})$. In the world of inequalities, the 'Difference Method' is our best friend.

We define a new function $D=P(C)-RHS$. If we can show $D\ge 0$, we have won. Substituting our expressions:

Since $z=x+y-xy$, we can substitute $x+y=z+xy$ into the expression for $D$:

Watch closely as the $z$ terms cancel out. We are left with:

The entire problem has collapsed into a single, elegant requirement: prove $z^2\ge xy$.

Remember that $z=P(A\cup B)$. Since $A\subseteq (A\cup B)$ and $B\subseteq (A\cup B)$, it must be true that $P(A)\le P(A\cup B)$ and $P(B)\le P(A\cup B)$. This gives us $x\le z$ and $y\le z$.

Since probabilities are non-negative, we can multiply these two inequalities:

This is the core of the proof. We have shown that $z^2-xy\ge 0$, which confirms that $P(C)\ge P(A\cup B)P(\bar{A}\cap \bar{B})$. You have navigated the logic and arrived at the truth.