Animated Solution for Mathematics - Probability: Let $E^c$ denote the complement of an event $E$. Let $E,F,G$ be pairwise independent events with $P(G)>0$ and $P(E\cap F\cap G)=0$. Then $P(E^c\cap F^c|G)$ equals

We are tasked with finding the probability of the intersection of the complements of two events, $E$ and $F$, given that event $G$ has occurred. The expression to evaluate is $P(E^c\cap F^c|G)$.

Our first step is to simplify the expression using De Morgan's Law. This law allows us to rewrite the intersection of complements as the complement of the union:

Consequently, our target probability transforms into:

Next, we apply the fundamental property of conditional probability, which states that the probability of the complement of an event is one minus the probability of the event itself. This yields:

To expand $P(E\cup F|G)$, we utilize the Inclusion-Exclusion Principle:

We are given that $E,F,$ and $G$ are pairwise independent. This implies that the occurrence of $G$ provides no information about $E$ or $F$. Mathematically, this simplifies our terms:

Substituting these into our previous expansion, the expression becomes:

Finally, we evaluate the term $P(E\cap F|G)$. By the definition of conditional probability:

Since the problem states that $P(E\cap F\cap G)=0$ and $P(G)>0$, this term vanishes. We are left with:

Recognizing that $1-P(E)=P(E^c)$, the final, elegant result is: