A Conditional Probability Question- Card Problem

Question

Here's a basic conditional probability question:

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?

So, let's say O denotes a side is orange and B denotes a side is black. Here's my attempt: $P(\text {O|other side B} )=\frac{P(\text {B|other side O} )P(\text{O})}{P(\text {B|other side O} )P(\text{O})+P(\text {B|other side B} )P(\text{B})}=\frac{1\times \frac{1}{4}}{1\times \frac{1}{4}+\frac{1}{2}\times \frac{3}{4}}=\frac{2}{5}$

But when I try in this way: $P(\text {O|other side B} )=P(O\cap B)/P(B)=1/3$

The answer is $\frac{1}{3}$, where is the mistake? Any help is appreciated. Thanks in advance.

Answer 1

Call the card with one orange and one black side the $OB$ card, and call the card with two black sides the $BB$ card.

$$P(\text{You picked $OB$ } \mid \text{you saw black side}) = \frac{P(\text{one side is black} \cap \text{ you picked OB})}{P(\text{you saw black side})} $$

$$= \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1} = \boxed{1/3}$$

In order to compute $P(\text{you saw black side})$, you need to condition on which card you drew. With probability $1/2$, you will pick $OB$ in which case you will see the black side with probability $1/2$. On the other hand, with probability $1/2$ you will pick $BB$ in which case you will see the black side with probability $1$. You must consider both cases.

Answer 2

The two card have in total 3 black sides Hence n(s)=3 Of which only one is favourable for other side to be orange , hence n(E) =1 P(E)=1/3

Answer 3

Alternative solution of the problem posed in your question.

Originally there are $6$ sides with equal probability to become the side viewed.

Then the condition is posed that the side viewed is black.

That means that $3$ candidates fall off and there are now $3$ sides left, again with equal probability to become the side viewed.

For exactly $1$ of these sides it is true that the other side is orange.

So the probability on that event equals $\frac13$.