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...

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.

• what if the weather forecast always predicts good weather right, but just sometimes fails to acknowledge that there will be rain? In that case, the overall correctness of the forecast will still be 2/3, but the propabilities you calculated are different... Apr 30, 2014 at 0:02
• That's a good point, the question should be more specific about the correlation of random variables. Given no other information I think you have to assume they're independently distributed, though. Apr 30, 2014 at 5:02

To get caught in the rain without an umbrella, the weather must rain, and the forecast must be wrong.

$$\frac12\text{rain}\times\frac13\text{wrong}=\frac16$$

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.

• I agree with your point about the fact good or bad weather might be better predicted etc..but where did you get those numbers for P(Rain and forecast rain) and P(No rain and forecast no rain) then?
– nx__
Apr 30, 2014 at 0:27
• they are just an example. any 2 numbers that sum up to $\frac 2 3$ would fit. They can be ($\frac 1 6$, $\frac 1 2$). When we assume the variables to be independent, as everybody else seems to do automatically, they are ($\frac 1 3$, $\frac 1 3$) and the result is ($\frac 1 6$, $\frac 1 6$). Or the numbers can be ($\frac 1 2$, $\frac 1 6$) which would be the other border case. Everything in-between is possible too, of course. Apr 30, 2014 at 8:15
• Anyhow, if we want to calculate P(Rain and forecast no rain) we have to make some assumptions. It is reasonable to assume that the variables are independent. If this is a homework assignment and I present the solution with dependent variables, the teacher would maybe just ask how the case would be for independent variables, and everything would be okay. But if you present a solution with independent variables and the teacher asks you about your extra assumptions, you may have a problem if you didn't think about dependent variables. Apr 30, 2014 at 8:21