It rains half the time in London....conditional probability It rains half the time in London and the weather forecast is correct in its prediction 2/3 of the time. You live in London and you always take an umbrella if the forecast predicts rain, and you never take an umbrella if the forecast does not predict rain. Calculate:
A) the probability that you are caught in the rain without an umbrella
B) the probability that you carried an umbrella on a day without rain
I used P(Rain) = P(Rain| Forecast Rain) * P(Forecast Rain) + P(Rain| Forecast no rain) * P(Forecast no rain).
Eventually I got P(Forecast Rain) = 1/2 and for A) I got 1/6 as an answer. Is this correct? I am pretty confused...
 A: Here's one way:
A) Caught in the rain without an umbrella is the event (Rain and Forecast No Rain). Now, 
$$P(\text{Rain and Forecast No Rain}) = P(\text{Forecast No Rain}|\text{Rain})\cdot P(\text{Rain}) = \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{6}$$
B) Carried an umbrella on a day without rain is the event (No Rain and Forecast Rain).
$$P(\text{No Rain and Forecast Rain}) = P(\text{Forecast Rain}|\text{No Rain})\cdot P(\text{No Rain}) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$
When in doubt with these kinds of problems, one thing you can always do is try to compute the conditional probabilities of the most specific events, ie, the 'and' events. Any more complicated event can be written as a disjoint union of the 'and' events, which is a sum for probabilities. So here, the specific events are (Rain and Forecast Rain), (Rain and Forecast No Rain), etc.
A: To get caught in the rain without an umbrella, the weather must rain, and the forecast must be wrong.
$$\frac12\text{rain}\times\frac13\text{wrong}=\frac16$$
A: The key point here is that we tend to assume that the validity of the forecast is independent of the weather. This does not need to be the case!
We don't know if the validity of the forecast is independent of the weather. The $\frac 2 3$ you mention mean that P(Rain and Forecast rain) + P(No rain and Forecast no rain) = $\frac 2 3$. There are many ways to solve this equation. One would be
P(Rain and forecast rain) = $\frac 1 6$
P(No rain and forecast no rain) = $\frac 1 2$
In this case the probability that you are caught in the rain without an umbrella is $\frac 1 2 - \frac 1 6 = \frac 1 3$. The probability that you carried an umbrella and although it doesn't rain is 0.
The general case is like this:
P(Rain and forecast rain) = $x$
P(No rain and forecast no rain = $y$
We know $x+y = \frac 2 3$.
P(caught in the rain without umbrella) = $\frac 1 2 - x$. Proof: P(Rain and forecast rain) + P(Rain and forecast no rain) = $\frac 1 2$...
P(carried umbrella but no rain) = $\frac 1 2 - y$. Proof: analogous.
