Is conditional probability always meaningful

Question

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?

Answer 1

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

$$Pr(B\mid A)=\frac{Pr(A\cap B)}{Pr(A)}=\frac{\frac{20}{72}}{\frac{40}{72}}=\frac12.$$

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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.

Answer 2

$P(A \mid B) \neq P(A)$.

$$P(A \mid B)= \frac{P(A \cap B)}{P(B)} = \frac{\frac5{18}}{\frac59\times\frac48+\frac49\times\frac38} = \frac{5}{8}$$

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 $P(A \mid B) = P(A)$.

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Conditional probability is always meaningful. $P(A \mid B)=P(A)$ also give information. It namely tells you that $A$ and $B$ are independent.

Answer 3

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.

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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...

Answer 4

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.