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The answer is the Probability that the first ball is red and the second ball is black, divided by the Probability that the second ballis 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,
$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$ = $\frac{\frac{7}{60}}{\frac{1}{4}} = \frac{7}{15}$
