Disprove $P(A\mid B) + P(A\mid\bar{B}) = 1$

Question

How can I disprove this statement?

$$P(A\mid B) + P(A\mid \bar{B}) = 1$$

I tried giving a counterexample but I'm asked to get to some kind of contradiction using sets $A$ and $B$.

Answer 1

Suppose that $P(B) = \frac{1}{2}$ and $P(A) = \frac{1}{2}$. Then, $P(\overline{B}) = 1-P(B) = 1 - \frac{1}{2} = \frac{1}{2}.$

Moreover, using the law of total probability:

$$P(A) = P(B)P(A|B) + P(\overline{B})P(A|\overline{B}) \Rightarrow $$ $$\frac{1}{2} = \frac{1}{2}P(A|B) + \frac{1}{2}P(A|\overline{B}) \Rightarrow $$ $$P(A|B) + P(A|\overline{B}) = 1.$$

As a consequence, you cannot disprove that

$$P(A|B) + P(A|\overline{B}) = 1$$

since there are random variables which satisfy the previous equation.

Answer 2

Take $A$ and $B$ such that $$P(A\cap B)\neq P(A)P(B).$$

For example, $P(A)=0.5$, $P(B)=0.3$ and $P(A\cap B)=0.1$.

Thus, $$\frac{P(A\cap B)}{P(B)}+\frac{P(A\cap\bar{B})}{P(\bar{B})}=\frac{1}{3}+\frac{4}{7}\neq1.$$$$

Answer 3

Take $A=\Omega$ then $P(\Omega|B)=1$ and $P(\Omega|\bar B)=1$.