I can solve this problem. An urn contains $4$ white and $5$ black balls; a second urn contains $5$ white and $4$ black ones. One ball is transferred from the first to the second urn. What is the probability that a ball then drawn from the second urn will be white?
Let $T_W$ represent the transferring of a white ball to the second urn while $T_b$ transfers a black one. Let $W$ represent the extraction of a white ball. Then the law of total probability yields
\begin{align*} P(B) &= P(B|T_B) P(T_B) + P(B|T_P) P(T_P)\\ &= \frac{6}{10} \frac{4}{9} + \frac{5}{10} \frac{5}{9}\\ &= \frac{24}{90} + \frac{25}{90}\\ &= \frac{49}{90}. \end{align*}
This problem is very clear to me. How can I adjust my knowledge to be able to solve the following one?
An urn contains $3$ white and $4$ black balls; a second urn contains $4$ white and $3$ black ones. (In extractions, the ball always goes back to the same urn where it came from --- id est, with replacement.) Extract a ball from a randomly chosen urn. If the ball is white, choose the next from urn $1$. If it's black, choose the next from urn $2$. Given that the first ball came from urn $1$, what's the probability that the second one is white?