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Suppose $A$ and $B$ are independent. Show that $A^{c}$ and $B$ are independent.

So $$P(A^{c} \cap B) = P(B)-P(B \cap A) $$ $$= P(B)-P(B)P(A)$$ $$= P(B)[1-P(A)]$$ $$= P(B)P(A^{c})$$

Is this right? I couldn't write it as $P(A^{c} \cap B) = P(A^{c})-P(A^{c} \cap B^{c})$ because we do not know that $A^{c}$ and $B^{c}$ are independent yet.

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I see no issues with your proof. – user21436 Jan 23 '12 at 15:31
Yes, it is right. It is possible that someone might expect justification of the line $P(A^c\cap B)=P(B)-P(B\cap A)$. I wouldn't. – André Nicolas Jan 23 '12 at 15:34

Just to elaborate on your observation:

You could proceed from the following step too, but that will be a rather unnecessarily complicated way of doing. Look at the following:

$$P(A^c \cap B)=P(A^c)-P(A^c\cap B^c)$$

Now use the De-Morgan's Law: $A^c \cap B^c=(A \cup B)^c$

So, what does that mean, $$\begin{align*}P((A \cup B)^c)&=1-P(A \cup B)\\&=1-P(A)-P(B)+P(A \cap B)\\&\overset{ind}{=}1-P(A)-P(B)+P(A) \cdot P(B)\\&=(1-P(A))\cdot(1-P(B))\\&=P(A^c)\cdot P(B^c)\end{align*}$$

Now, you have proved on the way that $A^c$ and $B^c$ are independent iff (try to prove one of the direction I haven't proved!) $A$ and $B$ are independent.

Now to complete, put this in there, $$\begin{align*}P(A^c \cap B)&=P(A^c)-P(A^c) \cdot P(B^c)\\&=P(A^c)[1-P(B^c)]\\&=P(A^c) \cdot P(B)\end{align*}$$

Hope you're convinced!

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