You have a hat in which there are three pancakes: One is golden on both sides, one is brown on both sides, and one is golden on one side and brown on the other. You withdraw one pancake, look at one side, and see that it is brown. What is the probability that the other side is brown?
Let's approach this problem step by step using conditional probability.
Given:
Let's call the event of observing a brown side on the withdrawn pancake as B.
$$ P(BB|B) = \frac{P(BB \cap B)}{ P(B)}. $$
To find $P(BB \cap B)$, we need to consider the probability of selecting the BB pancake and observing a brown side.
$$ P(BB \cap B) = 1/3 \cdot 1 = 1/3. $$
To find P(B), we need to consider all the ways we can observe a brown side.
$$ P(B) = P(BB \cap B) + P(GB \cap B),
$$
But, $P(GB \cap B) = 1/3 \cdot 1/2 = 1/6$ , and, $P(B) = 1/3 + 1/6 = 1/2$.
Therefore, the probability that the other side is brown, given that one side is brown, is:
$$ P(BB|B) = (1/3) / (1/2) = 2/3. $$