Prove $P(A|B^c)=P(A)$ 
Given that $A$ and $B$ are independent, Prove $P(A|B^c)=P(A)$

$P(A|B^c)=P(A\cap B^c)/P(B^c)=P(A)$ if $A$ is independent to $B^c$, which is not given.
Am I wrong?
 A: Well, then you will need to show that $A$ and $B^{c}$ are independent if $A$ and $B$ are.
So we need to show $P(A \cap B^{c}) = P(A)P(B^{c})$.
Let's do this using other equations we already know that involve some of the above terms:


*

*$P(A) = P(A \cap B) + P(A \cap B^{c})$, right?

*Also, since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$.
So substituting 2. into 1. gives us $P(A) = P(A)P(B) + P(A \cap B^{c})$, or rearranged: $$P(A) - P(A)P(B) = P(A \cap B^{c}).$$
Now, if only the left hand side (LHS) equaled $P(A)P(B^{c})$, we'd be done!  
But wait:
$$\text{LHS} = P(A) - P(A)P(B) = P(A)(1 - P(B))$$ and since $1 - P(B) = P(B^{c})$, we do get that the left hand side equals $P(A)P(B^{c})$.
So, $P(A \cap B^{c}) = P(A)P(B^{c})$, which means $A$ and $B^{c}$ are independent if $A$ and $B$ are.
A: Remember that
$$P(A) = P(B)P(A|B)+P(B^c)P(A|B^c).$$
Now, using the assumption that $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$, we get $P(A|B) = P(A \cap B)/P(B) = P(A)$. Now, solving the first equation for $P(A|B^c)P(B^c)$, we get
$$P(A|B^c)P(B^c) = P(A)-P(A)P(B) = P(A)(1-P(B)) = P(A)P(B^c),$$
from which we can solve $P(A|B^c) = P(A)$, which is what we wanted.
