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Probability theory! I try and try some more with this problem, and I just can't seem to get my head around it.

A weather forecast app can forecast 4 kinds of weather—rainy, sunny, snowy and cloudy. The accuracy of rainy forecast is 0.8, while the accuracy of sunny, snowy and cloudy forecast is 0.9. In the past 5 years, Paris had 10 percent rainy days. If the app shows that tomorrow is a rainy day, what is the probability that it is not going to rain?

I use bayes and the law of total probability, which implies:

$ p(NoRain|Rainyforecast) = \frac{p(Rainyforecast|NoRain)p(NoRain)}{p(Rainyforecast|NoRain)p(NoRain)+p(Rainyforecast|Rain)p(Rain)} $ How do I find the $p(Rainyforecast|NoRain),p(Rainyforecast|Rain)$? I can kind off deduce, that I have to relate them to the other forecast models, but I don't quite see how.

Thanks!

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    "The accuracy of rainy forecast is 0.8". Hence $P(RF|R) = 0.8$ and $P(RF|NR) = 0.2$. – 2017-02-06
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    No, my correction sheet says otherwise. Besides I don't see how a 0.8 accuracy can be interpreted as p(RF|R) = 0.8. It's the other way around p(R|RF) = 0.8. – 2017-02-06
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    That would make things easy since $p(NR|RF)=1-p(R|RF)$ and thus it would be equal to $0.2$. – 2017-02-06
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    But it's not. My solution sheet says p(RF|NR) = 0.1 – 2017-02-06
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    By the logic in your first comment, $P(NR|RF) = 1- .8 = .2$ and the problem is over. Which is what Shinja is also saying. I don't see how P(RF|NR) = .1 makes any sense. But the question is a bit confusing. I think your solution sheet might be wrong. – 2017-02-06
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    I really doubt the solution sheet is wrong. It's a big exam at one of germanies best technical universities. I think you're mixing the notations around, we're not interested in P(NR|RF), but P(RF|NR). – 2017-02-06
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    @1233023 You want the probability that it is *not rainy* given the app *forecast rain*. That *is* $\mathsf P(NR\mid RF) ~=~ 1 - \mathsf P(R\mid RF) ~=~ 0.2$. – 2017-02-07

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