<s>[INST] 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. 
 [/INST] 
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,
$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$ = $\frac{\frac{7}{60}}{\frac{1}{4}} = \frac{7}{15}$
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