Conditional Probability I have a question on conditional probability, which goes as follows: 
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
(b) the probability that he brings his umbrella, given that it doesn’t rain. 
I cannot seem to understand why the probability that he brings his umbrella is $\frac{5}{12}$. I seem to get $\frac{5}{9}$ Using the total probability theorem given that we are told it rains.
Please help and thanks in advance.
 A: Let $p_{ab}$ be the probability for $a,b\in\{r,n\}$ (for "rain" and "no rain"), with $a$ the forecast and $b$ the actual outcome. Then $p_{rr}=p_{nn}=\frac13$ and $p_{rn}=p_{nr}=\frac16$. Given that it doesn't rain, we're left with $p_{nn}=\frac13$ and $p_{rn}=\frac16$, so with probability $\frac23$ no rain is forecast and with probability $\frac13$ rain is forecast. In the former case Mr. Pickwick brings his umbrella with probability $\frac13$, and in the latter case with probability $1$. Thus he brings it with probability
$$
\frac23\cdot\frac13+\frac13\cdot1=\frac59\;,
$$
as you correctly calculated.
A simpler way to derive the probabilities $\frac13$ and $\frac23$ for the prediction is to use the symmetry between the prediction and the outcome: We can consider the actual outcome to be a "prediction" of the prediction, which is "correct" precisely if the actual prediction is correct, and thus with the same probabilities.
A: The answer 5/12 is for another question: the probability P(It does not rains, given the he brings his umbrella). Some books edition puts answers for changed exercises. Then P(not rain|brings umbrella) = (5/9 * 1/2) / (12/18) = 5/12 !!!
