We have two urns A and B. Urn A has 4 red balls and 3 blue balls, and urn B is empty. Randomly draw two balls $X_1$ and $X_2$ from urn A without replacement and place them into urn B. Then, draw two balls $Y_1$ and $Y_2$ from urn B with replacement.
Given that $Y_1$ and $Y_2$ are both blue, what is the probability that $X_1$ and $X_2$ are different colors?
So to start this problem, I labeled the event "$X_1$ and $X_2$ are different colors" as event A, and labeled the event "$Y_1$ and $Y_2$ are both blue" as event B. So what I want to find is $P(A \mid B)$.
Using the definition of conditional probability, we have that $$P(A \mid B)= \frac{P(A \cap B)}{P(B)}$$ and $$P(A \cap B)=P(B \mid A)P(A)$$ I computed $P(B \mid A)=\frac{1}{4}$ and $P(A)=\frac{4}{7}$, giving me $P(A \cap B)=\frac{1}{7}.$
Then using the law of total probability, we have that $$P(B)=P(B \mid A)P(A)+P(B \mid A^c)P(A^c)$$ which is where I am running into trouble. How can I calculate $P(B \mid A^c)$, or in plain English, the probability that $Y_1$ and $Y_2$ are both blue given that $X_1$ and $X_2$ are the same?
There are $\binom42=6$ ways to draw two red balls from urn $1$ and $\binom32=3$ ways to draw two blue balls. So the probability that both balls are blue, given that they are the same color, is $$\frac3{6+3}=\frac13$$