Probability of the backside being black If I have $n$ cards where one side is white and the other is black and $m$ cards where both sides are black and I draw a random card, what is the probability of the backside of the card being black given the condition that the frontside is black? My approach: Let $A$ be the event where the backside is black and $B$ the event where the frontside is black. $P(A| B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{2m+n}{2(m+n)}\cdot \frac{2m+n-1}{2(m+n)-1}}{\frac{2m+n}{2(m+n)}}=\frac{2m+n-1}{2(m+n)-1}$ Is that correct? I feel like I didn't get it right.

The probability of drawing a card with two different sides equals $\frac{n}{m+n}$, and the probability of seeing the black side is $\frac{1}{2}$. The probability of drawing a card with two black sides is $\frac{m}{m+n}$, with a probability of $1$ to see a black side. As such, we get:

$$P[A|B] = \frac{P[A,B]}{P[B]} = \frac{\frac{m}{m+n}}{\frac{n}{m+n} \cdot \frac{1}{2} + \frac{m}{m+n}} = \frac{2m}{2m+n}$$

You can also look at it this way: there are $2m+n$ black sides, of which $2m$ have a black partner. As such, the probability equals:

$$\frac{2m}{2m+n}$$