n cards(one side white, one black), m cards(both sides black)

Question

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.

Answer 1

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.

Answer 2

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: $$P(A|B)=\frac{\frac{m}{m+n}}{\frac{2m+n}{2m+2n}}=\frac{2m}{2m+n}$$