$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=1$ is false in general

Question

Let $A,B$ be some events in probability space $\Omega$. Show that $$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=1$$ is false in general.

I thought of giving an example there events $A$ and $B$ are independent but then I get the correct equation because $$\mathbb{P}(A|B)+\mathbb{P}(A^c|B^c)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}+\frac{\mathbb{P}(A^c \cap B^c)}{\mathbb{P}(B^c)}=\mathbb{P}(A) + \mathbb{P}(A^c)=1.$$

My next idea was to check all other cases like when $A \subset B$ or when $A=B$ but everything seems to work out well. Can you help me of thinking of an example?

Answer 1

$$P(A|B)+P(A^c|B^c)=1\\\iff \frac c{c+d}+\frac a{a+b}=1\\\iff 2ac+bc+ad=ac+ad++bc+bd\\\iff ac=bd\\\iff A \text{ and } B \text{ are }\textbf{independent events}.$$

The last equivalence is from my previous derivation here.

Therefore, the given statement is not in general true.

P.S. To say that the given statement is “false in general” may suggest that it cannot be true, which is not quite correct.

P.P.S. I wrote this related Answer, to another Question asked around the same time.

Answer 2

What is true is $$P(A\mid B)+P(A^c\mid B)=1.$$

If your equality was true, it would mean $$P(A^c\mid B)=P(A^c\mid B^c),$$ which is the definition of

$A^c$ and $B$ are independent.

That can be shown to be equivalent to

$A$ and $B$ are independent.

So your equality is true if and only if $A$ and $B$ are independent.

But for example, if $A$ and $B$ are the same event, then $$P(A\mid A)+P(A^c\mid A^c)=2.$$

Then your equality is not true.

If $B=A^c$ then you get $$P(A\mid A^c)+P(A^c\mid A)=0.$$

So your sum can really be anywhere between $0$ and $2.2

Answer 3

You could let $A$ be any event with probability strictly between $0$ and $1$, and let $B=A^c$.

Then $P(A|B)+P(A^c|B^c)=P(A|A^c)+P(A^c|A)=0$