drawing balls from an urn (conditional probability) Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is  randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black?
I encountered this problem in my AoPS textbook, as a complete newcomer to conditional probability. Is the answer 1/2? I would really appreciate a solution, as I have no way to check my work. 


The answer is the Probability that the first ball is red and the second ball is black, divided by the Probability that the second ball
is black.
$\frac{P(\text{First Ball is RED AND  Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$
P(1st ball = Red and 2nd ball = Black  )
$=\frac{1}{2}[ (\frac{2}{6}* \frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) ] = \frac{7}{60}$
P(Second ball is BLACK) = $\frac{1}{2}[ (Urn A's Redball*\frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) + ( \frac{3}{6}* \frac{1}{2}*\frac{2}{5} ) ] = \frac{1}{2}[ (1*\frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) + ( \frac{3}{6}* \frac{1}{2}*\frac{2}{5} ) ] = \frac{1}{4}$
Therefore,
$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$ = $\frac{\frac{7}{60}}{\frac{1}{4}} = \frac{7}{15}$
