Carying an umbrella given rain forecast We know it rains half of the time. 
We also know that the weather forecaster is correct 2/3 of the time. I.e. the probability that it rains given they had predicted rain is the same as if it does not rain given they predicted it would not rain and is 2/3.
We know a man takes his umbrella when rain is in the forecast and takes his umbrella with probability 1/3 when rain is not in the forecast.
I have read other threads on here with this problem and I have two answers which i think are correct; however, the answer key in my book only agrees with one answer. The question asks us to find the probability that: a) the man has no umbrella given that it rains. Then b) the probability that he brings his umbrella given it does not rain.
I made this tree

This is how I attempted to solve this.
Let A be the event that the man has his umbrella and r be the event that it rains.
first we want the probability the man has no umbrella given that it rains
$$P(A^c|r)=\frac{P(A^c \cap r)}{P(r)}=\frac{\frac{1}{2}\frac{2}{3}\frac{1}{3}}{\frac{1}{2}} = \frac{2}{9}$$
This answer matches the one in my answer key. The next one is the one that is off. My answer key says it should be $\frac{5}{12}$ which we will see is not what i get.
the probability that he brings his umbrella given it does not rain.
$$P(A|r^c)=\frac{P(A \cap r^c)}{P(r^c)}=\frac{\frac{1}{2}(1)\frac{1}{3}+\frac{1}{2}\frac{1}{3}\frac{2}{3}}{\frac{1}{2}} = \frac{5}{9}$$
Have I done something wrong? I am beginning to think the answer key is off.
 A: You have most ideas right.  This is just a correction because the root of the tree has not been given (unconditional probability of a rain forecast).  But posting it here because it's too large for the comments area.
Let p be the unconditional probability of a forecast of rain(F).  In 2/3 of those cases it rains(R). Denote the events no-rain and forecast of no-rain by R' and F' respectively.  Since you know $P(R|F)=2/3$ and $P(R|F')=2/3$, you can solve for $p$ because $P(R)$ is given to be 1/2.
$$P(R)=P(F)P(R|F) + P(F')P(R|F')=p\frac{2}{3}+(1-p)\frac{1}{3}=\frac{1}{2}$$ from which we know $p=\frac{1}{2}$. 
From here, you can also calculate $P(F|R')=\frac{1}{3}$ and $P(F'|R')=\frac{2}{3}$.  
So you can construct the tree directly relevant for each part of the question.  For part 2, start with the event R', which branches off to F|R' and F'|R' which were obtained just above. From each, there are further two nodes (whether or not he carries the umbrella).  
$$P(U|R')=P(U|F')P(F'|R') + P(U|F)P(F|R')=\frac{1}{3}\times\frac{2}{3}+1 \times \frac{1}{3}=\frac{5}{9}$$
