<s>[INST] When do we expand the numerator of the Bayes' Theorem I am trying to understand why the proposed solution below to the following question is wrong:-
A box contains three cards: a card that is black on both sides, one that is white on both sides and a card that is black on one side and white on the other. You withdraw a card at random and see that the visible face is black. What is the probability that the other side of this card is also black? 
Proposed solution: 
Let A = {other side is Black} and B = {observed side is Black}, then 
$$ P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ where,
$$ P(B|A) = \frac{1}{2}$$
$$ P(A) = \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot 0 = \frac{1}{2}$$
$$P(B) = P(B|A)P(A) + P(B|\tilde{A}) P(\tilde{A})$$ 
$$ = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot \frac{1}{2}= \frac{1}{2}$$
where $\tilde{A}$ ={other side is not Black}. Therefore,
$$ P(A|B) = \frac{\frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}} = \frac{1}{2} \textbf{  (wrong!)}$$
The proposed solution is incorrect. The correct solution is as follows:
$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
where,
$$ P(A\cap B) = \frac{1}{3}$$
$$ P(B) = P(B|C_{B}) P(C_{B}) + P(B|C_{W}) P(C_{W}) + P(B|C_{M}) P(C_{M})$$
$$ =\frac{1}{3}\cdot1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot\frac{1}{2} = \frac{1}{2}$$
where $C_{B}$ = {Card with both sides Black}, $C_{W}$ = {Card with both sides White} and $C_{M}$ = {Card with Mixed sides}. This gives,
$$P(A|B) = \frac{2}{3} \textbf{ (correct!)}$$ 
It seems that expanding the numerator $P(A \cap B)$ as $P(B|A) P(A)$ gives the wrong solution. Where is the proposed solution going wrong?
 [/INST] Your error is in asserting that : $\mathsf P(B\mid A) = \tfrac 1 2$   It does not.
The probability that the observed side is black given that the other side is black is $2/3$.   Of the six (equally probable) sides that the unobserved side could be, three of them are black.   Of these three black sides, only two of them have a black side on the flip-side of their card.
Then $\qquad\mathsf P(A\mid B) = \dfrac{\mathsf P(B\mid A)\mathsf P(A)}{\mathsf P(B)} = \dfrac{\tfrac 2 3\cdot\tfrac 1 2}{\tfrac 1 2} = \dfrac 2 3$
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