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In order to apply the Bayes-theorem to a real world example, I have been given this problem :

A barometer is used to forecast the weather. However the prediction may be erroneous. It is observed that in $20$ cases over $200$ rainy days the barometer has predicted good weather, and in $20$ cases over $100$ good days it has predicted rain. The local tourist guide says that $10\%$ of the days in a year are rainy. What is the probability that it will rain if the barometer predicts rain ?

Answer $\approx 0.333 $

My attempt : Let $R$ be the event "it rains" and $B$ the event "the barometer predicts rain". I am asked to compute $P(R|B)$. By the bayes-theorem, I know that $$P(R|B) = \frac{P(B|R)P(R)}{P(B)} = \frac{P(B|R)P(R)}{P(B\cap R) + P(B\cap \bar{R})} = \frac{P(B|R)P(R)}{P(B|R)P(R) + P(B|\bar{R})P(\bar{R})}$$ If I am not mistaken, I am given that $P(R) \approx 1/10$, $P(R \cap \bar{B}) \approx \frac{20}{200} = 1/10$ and $P(\bar{R} \cap B) \approx \frac{20}{100} =1/5$.

My question : How can I calculate the remaining unknown $P(B|R)$ (or $P(B \cap R)$)?

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    Hint: it's given in the text. $\mathrm{P}(B | R)$ is the probability that the barometer predicts rain on a rainy day, which would be the given fraction of rainy days where the barometer actually predicts rain. – 2017-01-10
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    I don't know how I missed that :d Thanks! – 2017-01-10

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You are close, but not quite correct. It is not true that $\Pr[R \cap \bar B] = 1/10$ and $\Pr[\bar R \cap B] = 1/5$. These are probabilities of the joint event occurring. What you were instead supplied are conditional probabilities: for example, in the first case, when the weather actually did rain $200$ times, the barometer was in error $20$ of these times. When it actually did not rain $100$ times, the barometer was in error $20$ times. This would suggest that $$\Pr[\bar B \mid R] = \frac{20}{200} = \frac{1}{10}, \\ \Pr[B \mid \bar R] = \frac{20}{100} = \frac{1}{5},$$ respectively. Then to recover the respective joint probabilities, you must multiply by the probability of the given event (i.e. $\Pr[R]$ the unconditional probability of rain, and $\Pr[\bar R]$, the unconditional probability of no rain). This is because it is much more likely that there is no rain; thus the respective probability of an erroneous prediction in the case of no rain is given greater weight than the probability of an erroneous prediction in the case of rain.

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    How can I convince myself that the problem does not give me the joint probability? I feel like the problem is ambiguous in the sense I could choose either to interpret it by "It will rain if the barometer predicts rain in 90% of the cases" and "It will rain and the barometer predicts rain in 90% of the cases" depending on weither you measured it before it happens or not. – 2017-01-10
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    You know that it must be a conditional probability that is being described, because you are told that in the $200$ times that it rains, the barometer is wrong $20$ times. In other words, **given** that it rained, the barometer was wrong $10\%$ of the time. It does not mean that the probability of it raining *and* the barometer being wrong is $1/10$, since it only rains on average once out of $10$ days. – 2017-01-10
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    Another way to understand that it must be conditional is to observe that the probability of rain on any given day is $\Pr[R] = 0.10$. Therefore, it is nonsensical to say that $\Pr[B \cap R] = 0.90$. The chance $B$ and $R$ occurring simultaneously cannot be greater than the probability of $R$ occurring irrespective of event $B$. – 2017-01-10
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    Makes sense! Thanks :) – 2017-01-10