Question :Is the following statement right or wrong?
If $P(A\mid B)=P(A\mid B^c)\rightarrow \:A$ and $B$ :aren't collectively exhaustive events.
Where $A$ and $B$ are independent and $P(A)>0$ , $P(B)<1$
What I did
I started supposing that the statement is false, so
if $A\cup B=Ω,\:\:B^c=A$ in that case
$P(A\mid B)=P(A\mid B^c)$
$P(A\mid B)=P(A\mid A)$ Because $B^c=A$
$P(A\mid B)=P(A\mid A)$
$P(A)=\frac{P(A\cap A)}{P(A)}$ $A$ and $B$ are independent
$P(A)=\frac{P(A)}{P(A)}$
$P(A\mid B)=1$ or $P(A)=1$
My question is does this result makes sense, implies that is false? Can $P(A)=1$ be right? because that would imply that $P(A)=P(Ω)$ right?
If sets $\ A,\ B\ $ are collectively exhaustive then $P(A \cup B) = 1.$
I don't see why you can't have $\ P(A)=1,\ B\subset A,\ $ and $P(B) = 0.$ This would fulfil all the requirements, right?
We would need $\ P(B)=0\ $ though, as if $\ P(A)=1\ $ and $\ P(B) > 0,\ $ then $P(B\vert A)>0=P(B\vert A^c),\ $ contradicting independence.
However, I think if we adjust the restriction on event $A$ to $\ P(A)<1,\ $ then we would have:
$(1-P(B))(1-P(A))>0 \implies P(A)+P(B)-P(A)P(B) < 1 $
$$\overset{\mbox{(indep)}}{\implies} P(A)+P(B)-P(A\cap B) < 1 $$
$$\implies P(A\cup B) < 1, $$
meaning $A$ and $B$ are not collectively exhaustive.