Bayes Probability - Rain and Umbrellas

I do not have the answers for this question, and was hoping if anyone could provide feedback on my thinking.

In London, half of the days have some rain. The weather forecaster

is correct 2/3 of the time, i.e., the probability that it rains, given that she has

predicted rain, and the probability that it does not rain, given that she has

predicted that it won’t rain, are both equal to 2/3. When rain is forecast,

Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with

probability 1/3. Find

(a) the probability that Pickwick has no umbrella, given that it rains.

If rain is predicted, regardless of the result - Pickwick will take the umbrella. So prob =0

If rain is not predicted and it rains and he does not have an umbrella = $\displaystyle \frac{1}{3}\times \frac{2}{3}=\frac{2}{9}$

(b) the probability that it doesn’t rain, given that he brings his umbrella.

When I used the tree diagram with a three step [rain(1/2)/not rain(1/2), prediction and result, umbrella/no umbrella] I am confused what the numerator is.

Thank you in advance for any assistance provided

Right, it's a harder problem than most realize

(Happy) This is a very favorite problem of mine.

Right, it’s not easy to see all the conditionals.

You actually have to do some algebra to get that the probability of her forecasting rain is ½ ,

and that the probability of her having forecasted rain given that it rains = the probability of her not having forecasted rain given that it does not rain = 2/3

Once this is done, you can easily fill in the tree.

See my website http://duport.com for a hint on the full solution.