Is conditional probability always meaningful Problem:
A bag contains $4$ red and $5$ white balls. Balls are drawn from the bag without replacement.
Let $A$ be the event that first ball drawn is white and let $B$ denote the event that the second ball drawn is red. Find 
(i) $P(B\mid A)$
(ii) $P(A\mid B)$
My confusion is that should $P(A\mid B)=P(A)$
Can we say that in general if $P(A\mid B)$ exists then $P(B\mid A)$ should also exist?
 A: Let the event space be
$$\Omega=\{(r,r),(r,w),(w,r),(w,w)\}$$
the corresponding probabilities are
$$\frac{12}{72},\frac{20}{72},\frac{20}{72},\frac{20}{72}.$$
Then
$$Pr(A\cap B)=Pr((w,r))=\frac{20}{72}$$
and
$$Pr(B)=Pr(\{(r,r),(w,r)\})=\frac{12}{72}+\frac{20}{72}=\frac{32}{72}$$
and
$$Pr(A)=Pr(\{(w,r),(w,w)\})=\frac{40}{72}$$
so,
$$Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{\frac{20}{72}}{\frac{32}{72}}=\frac58.$$
and
$$Pr(B\mid A)=\frac{Pr(A\cap B)}{Pr(A)}=\frac{\frac{20}{72}}{\frac{40}{72}}=\frac12.$$

In general we cannot say that if $Pr(A\mid B)$ exists then $Pr(B\mid A)$ also exists. Let simply $Pr(A)=0<Pr(B)$. Then $Pr(A\mid B)=0$ but $Pr(B\mid A)$ is not defined.
A: In general:
$$P(B|A)P(A)=P(A\cap B)=P(A|B)P(B)$$
If you can find $P(A),P(B)$ and $P(A\cap B)$ then this enables you to find $P(A|B)$ and $P(B|A)$.
Note that $P(A|B)=P(A)$ leads to $P(A\cap B)=P(A)P(B)$ i.e. independence of $A$ and $B$. 
In your question $A$ and $B$ are not independent.

Hints:


*

*To find $P(B)$ realize that the $9$ balls all have equal probability to become the second ball drawn, and $4$ of them are red, so...

*Actually finding $P(B|A)$ "directly" is somehow easyer than finding $P(A\cap B)$. If the first ball has been drawn and is white then there are $8$ balls left and $4$ of them are red, so...
A: $P(A \mid B) \neq P(A)$. 
$$P(A \mid B)= \frac{P(A \cap B)}{P(B)} = \frac{\frac5{18}}{\frac59\times\frac48+\frac49\times\frac38} = \frac{5}{8}$$
We have $P(A)=\frac{5}{9}$. The intuition behind this is that $B$ makes it more likely that a white ball has been drawn the first time, because then $B$ is more likely. 
The key in the problem is that the balls are drawn without replacement. Otherwise we would have $P(A \mid B) = P(A)$.

Conditional probability is always meaningful. $P(A \mid B)=P(A)$ also give information. It namely tells you that $A$ and $B$ are independent. 
A: 
Can we say that in general if $P(A\mid B)$ exists then $P(B\mid A)$
  should also exist?

Not necessarily. I modifiy your exercise.

A bag contains $4$ red and $5$ white balls. Balls are drawn from the
  bag without replacement.
Let $A$ be the event that first ball drawn is white and let $B$ denote the event that the second ball drawn is black. Find   
(i) $P(B \mid A)$
(ii) $P(A\mid B)$

$P(B)=0$, therefore $P(A\mid B)=\frac{P(A \cap B)}{P(B)}$ is not defined.
