n cards(one side white, one black), m cards(both sides black) We have $n$ cards that are black on one side and white on the other and $m$ cards that are black on both sides. 
We draw a card and we only see one side; this side is black.
What is the probability that the other side is black as well?
I tried to use the rule of Bayes.
Let $A=$ "both sides black", $B=$ "one side black, one white"
So $A=\frac{m}{n+m}$, $B=\frac{m}{n+m}$.
So $P(A\cap B)=\frac{nm}{(n+m)^2}$.
And so I get $\frac{P(A \cap B)}{P(B)}=\frac{n}{n+m}$.
I do not know what I am doing wrong. Probably my sets $A$ and $B$ are not correctly chosen. Help is much appreciated.
 A: Your events $A$ and $B$ are disjoint, so $P(A\cap B)=0$. In fact $B=\overline A$.
You need to consider the following events: $A=$ both sides black, $C=$ black side seen. Then $P(A)=\frac n{n+m}$ and $P(\overline A)=\frac m{n+m}$ as you say. If $A$ happens then certainly $C$ also happens, so $P(A\cap C)=\frac n{n+m}$. The probability you want is $P(A\mid C)$.
$P(C\mid A)=1$ and $P(C\mid\overline A)=\frac12$. Now use the relation
$$P(C)=P(A)P(C\mid A)+P(\overline A)P(C\mid\overline A)$$
to get $P(C)$, and then apply Bayes' rule.
A: Bayes Theorem states:
$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$
To apply this to your problem, we are given that one face is black, and we wish to know the probability that the card is type 1 (both sides black).
So let A be the event the card is type 1, and B be the event the face revealed is black.
$P(B|A)=1$, so we are left with:
$$P(A|B)=\frac{P(A)}{P(B)}$$
We know $P(A)=\dfrac{m}{m+n}$.
$P(B)$ can be calculated from the fact that there are $2(m+n)$ possible faces to be revealed, of which $2m+n$ are black.
So $P(B)=\dfrac{2m+n}{2m+2n}$
So:
$$P(A|B)=\frac{\frac{m}{m+n}}{\frac{2m+n}{2m+2n}}=\frac{2m}{2m+n}$$
