Suppose you have two cards: one is painted black on both sides and the other is painted black on one side and orange on the other You select a card at random and view one side.
You notice it is black. What is the probability the other side is orange?
What I have done is following:
Card 1: $B_{1}$, $B_{2}$
Card 2: $B_{1}$, $O_{2}$
Now, we want to calculate $P(O_{2}|B_{1})$, which is $$P(O_{2}|B_{1})=\dfrac{P(B_{1}\cap O_{2})}{P(B_{1})}$$
$P(B_{1}\cap O_{2})=\dfrac{1}{2}$ as there is only $1$ card (out of $2$) giving us black and orange. $P(B_{1})=\dfrac{3}{4}$.
Therefore, the resulting conditional probability is $\dfrac{2}{3}$
However, the solution of this question telling me $\dfrac{1}{3}$ is the correct answer (without explanation, just a number).
What's wrong here?
Thank you!
$P(B_1\cap O_2)=\frac14$, since to view a black side when the other side is orange requires first selecting the black/orange card and then viewing the black side on that card, both of which have a $\frac12$ chance of occurring.