# pairwise & joint independence of three events implies independence of events involving their complements

This question is for an intro class in probability theory :)

Let $$A, B, C$$ be three events that are pairwise & jointly independent. Show that $$A^c$$ and $$B \cup C^c$$ are also independent.

I started with some definitions. Pairwise & joint independence give us the following: $$P(A \cap B) = P(A)P(B)$$ $$P(B \cap C) = P(B)P(C)$$ $$P(A \cap C) = P(A)P(C)$$ $$P(A \cap B \cap C) = P(A)P(B)P(C)$$

I could also show that pairwise independence of the complements of the events holds, ie. $$P(A^c \cap B^c) = P(A^c)P(B^c)$$ $$P(B^c \cap C^c) = P(B^c)P(C^c)$$ $$P(A^c \cap C^c) = P(A^c)P(C^c)$$

The way I'm thinking about this problem is to simply follow the definition of independence. Show that: $$P(A^c \cap (B \cup C^c)) = P(A^c) \cdot P(B \cup C^c) = P(A^c) \cdot (P(B) + P(C^c) - P(B \cap C^c))$$ And the work I've done so far: $$P(A^c \cap (B \cup C^c)) = P((A^c \cap B) \cup (A^c \cap C^c)) = P(A^c \cap B) + P(A^c \cap C^c) - P((A^c \cap B) \cap (A^c \cap C^c) = P(A^c \cap B) + P(A^c \cap C^c) - P((A^c \cap (B \cap C^c)) = P(A^c)P(B) + P(A^c)P(C^c) - P(A^c \cap (B \cap C^c))$$ This is where I get stuck. I think this method reduces the problem to proving $$A^c, (B \cap C^c)$$ are independent, which I can't seem to do. Any hints?

$P(A^{c} \cap (B\cap C^{c}))=P (B\cap C^{c}))-P(A \cap (B\cap C^{c}))$. Now $P (B\cap C^{c}))=P(B)-P(B\cap C)=P(B)-P(B)P(C)$. Also $P(A \cap (B\cap C^{c}))=P(A\cap B) -P(A\cap B \cap C)=P(A)P(B)-P(A)P(B)P(C)$. Now just put all these together to get independence of $A^{c}$ and $B\cap C^{c}$.