If I can solve this simple urn probability problem, what do I need to solve this more complex urn problem?

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?

• Use exactly the same approach. what are the ways to get a white ball in the second draw? (i) the first draw is black and the second draw is white, or (ii) the first draw is white and the second draw is also white. What happens in the two cases? – onetimething Apr 15 '18 at 0:34
• @Onetimething Here's what I do, but I'm totally not sure. To begin there are two possibilities: I could draw from first urn or second urn. However, the problem tells me the first ball came from first urn, so I can completely ignore the part of the sample space that describes first draw from second urn. Then I get a new problem: what's the probability we can draw a white ball? Now I have two possibilities: if the first ball was white, I'll use the first urn; otherwise, second urn. Based on that, I set up the following. – Joep Awinita Apr 15 '18 at 15:50
• Let $B_1$ mean black ball was the first drawn; let $W_1$ mean white ball was the first drawn. Then $P(W) = P(W| W_1) P(W_1) + P(W|B_1)P(B_1) = \frac{3}{7}\frac{3}{7} + \frac{4}{7}\frac{4}{7} = \frac{25}{49}$. – Joep Awinita Apr 15 '18 at 15:50
• @Onetimething If this is right, there's a next problem for me. Same problem but I'd want the probability that the third ball is a white given the first ball came from urn 1. I have not seen a way to solve it yet. It must be a very similar reasoning, but I haven't spotted it yet. Any directions will be appreciated. – Joep Awinita Apr 15 '18 at 19:02
• Yes, this seems perfectly reasonable to me. For the case of three draws too as you mentioned use the same approach. But there will be a lot more terms to add/multiply! – onetimething Apr 15 '18 at 23:58