Animated Solution for Mathematics - Probability: If $E$ and $F$ are independent events such that $0<P(E)<1$ and $0<P(F)<1$, then

Welcome, fellow traveler in the realm of probability. Today, we are going to dissect a problem that seems simple on the surface but holds the very essence of what it means for two events to be 'independent.'

When we say $E$ and $F$ are independent, we are not just stating a formula; we are describing a world where the occurrence of one event is completely indifferent to the other. Let us embark on this journey to understand why this property is so powerful.

The Myth of Mutual Exclusivity

Imagine our sample space $S$ as a vast, abstract canvas. Inside this canvas, we have two regions, $E$ and $F$. We are told that $0<P(E)<1$ and $0<P(F)<1$.

This is our starting point—neither event is impossible, nor is either guaranteed. Now, consider Option A: are they mutually exclusive?

To be mutually exclusive means that $E$ and $F$ cannot coexist; their intersection must be empty, meaning $P(E\cap F)=0$. But we know that $E$ and $F$ are independent. By definition, independence dictates that $P(E\cap F)=P(E)P(F)$.

Since both $P(E)$ and $P(F)$ are strictly positive, their product must also be strictly positive. Thus, $P(E \cap F) eq 0$. The logic is ironclad: independent events with non-zero probabilities can never be mutually exclusive. They must overlap.

The Algebra of Complements

Now, let us tackle the more interesting questions: what happens when we look at the complements, $E^c$ and $F^c$?

Consider Option B: are $E$ and $F^c$ independent? To check this, we need to see if $P(E\cap F^c)=P(E)P(F^c)$.

Look at the Venn diagram. The region $E\cap F^c$ is simply the part of $E$ that does not overlap with $F$. Mathematically, we can write this as $P(E\cap F^c)=P(E)-P(E\cap F)$.

Substituting our independence condition, $P(E\cap F)=P(E)P(F)$, we get:

Factoring out $P(E)$, we find:

Since $1-P(F)=P(F^c)$, we arrive at $P(E\cap F^c)=P(E)P(F^c)$. The equality holds! $E$ and $F^c$ are indeed independent. This is a beautiful result—if two events are independent, their complements are also independent.

Following the same logic for Option C, we look at $E^c$ and $F^c$. Using De Morgan's Law, we know that $E^c\cap F^c=(E\cup F{)}^c$. Therefore, $P(E^c\cap F^c)=1-P(E\cup F)$.

Expanding the union, we have $P(E\cup F)=P(E)+P(F)-P(E)P(F)$. Substituting this back, we get:

Rearranging the terms, we get $(1-P(E))-P(F)(1-P(E))$, which factors perfectly into $(1-P(E))(1-P(F))$, or $P(E^c)P(F^c)$. The symmetry is breathtaking, isn't it?

The Universal Truth of Conditional Probability

Finally, we arrive at Option D: $P(E|F)+P(E^c|F)=1$. This is a fundamental identity.

By the definition of conditional probability, $P(E|F)=\frac{P(E\cap F)}{P(F)}$. Similarly, $P(E^c|F)=\frac{P(E^c\cap F)}{P(F)}$. Adding these together, we get:

Since $E$ and $E^c$ partition the sample space, their intersections with $F$ must partition $F$ itself. Thus, $P(E\cap F)+P(E^c\cap F)=P(F)$.

The expression simplifies to $\frac{P(F)}{P(F)}=1$. This holds true regardless of whether the events are independent or not!

We have navigated the landscape of probability, proving that independence is not just a condition, but a structural relationship that preserves itself through complements.

We have seen that while independent events cannot be mutually exclusive, they share a deep, algebraic harmony. Keep this logic in your toolkit, and you will find that even the most complex probability problems begin to unfold with elegance.