How to combine annual rainfall data with specific forecast of known accuracy to predict weather for a particular day?

Alice is getting married tomorrow at an outdoor ceremony in a beautiful garden.

In recent years, it has rained, on average, only 5 days in a year. Alice is anxious about the weather for tomorrow and wishes to get the weather prediction from a weather forecaster.

From the past records, the weather forecaster is correct 90% of the time. Unfortunately, the weather forecaster predicts rain for tomorrow.

Should alice make wet weather plans for tomorrow?

Found, e.g., at http://stattrek.com/probability/bayes-theorem.aspx.

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Another thing you'd want to know is which five days it rained on. –  Peter Flom Sep 14 '12 at 12:04
I downvoted this question because it was originally taken wholesale from another source without attribution. –  whuber Sep 19 '12 at 14:30
@whuber that is not true. There is a subtle difference between the question OP asked and the one on that website. This question only specifies the condition that the weatherman is correct 90% of time; the question on the website gives the probability P(weatherman says rain | rain) and P(weatherman says rain | no rain). Now, I think it is likely that the question was taken without attribution and then someone failed to accurately type it up, but as Henry has shown we can generate useful bounds despite the fact that we can't answer the question the OP asked precisely. –  guy Sep 19 '12 at 17:46
@guy, thanks for pointing that out: I did not read the other version carefully, once I had discovered it was a very close match. I agree with you that Henry has a useful answer (and the other respondents have contributed worthwhile replies too); that's one reason I have not voted to close the question altogether. However, I feel strongly--and this reflects site policy--that using the work of other people (even if it is a question or variation thereof) without attribution is wrong. Downvoting appears the best way to recognize that in this case. –  whuber Sep 19 '12 at 17:49

There really isn't sufficient information here. In fact, the weather forecaster could be right a lot more than 90% of the time just by saying it will never rain; if it only rains 5 days per year, then forecast "no rain" will be right 360/365 = 98.6% of the time.

What you want to know is, when the weatherman forecasts rain, what % of the time does it rain?

This is similar to problems of sensitivity and specificity, with a different setting.

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Thank you. I adjusted the question to "Should alice make wet weather plans for tomorrow ?". I copied this question directly from my CAs with nothing changed in the main body. The question is made up by me. –  kingboonz Sep 14 '12 at 10:00
That doesn't really help. –  Peter Flom Sep 14 '12 at 10:01
You can bound the probability –  Henry Sep 18 '12 at 6:58

You need to deconstruct "the weather forecaster is correct 90% of the time". Note that $\frac{5}{365} \approx 1.37\%$

For example the weather forecaster might predict rain $11.37\%$ of the time, always correctly when it in fact rains, so overall in error $11.37\% - 1.37\% = 10\%$. Then about $\frac{1.37\%}{11.37\%} \approx 12.05\%$ of rain forecasts will in fact be followed by rain.

Or the weather forecaster might predict rain $8.63\%$ of the time, never correctly when it in fact rains, so overall in error $8.63\% + 1.37\% =10\%$. Then no rain forecasts will in fact be followed by rain.

Any other pattern will be between these, so Alice can conclude that the probability of rain given a rain forecast is less than or equal to $12.05\%$.

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How can you rule out the possibility that the forecaster always correctly predicts rain, but is only 89.86% accurate on the other days? That gives 100%, not 12.05%, as an upper bound on the answer. –  whuber Sep 18 '12 at 16:38
@whuber: That is my second paragraph: if the forecaster predicts rain correctly 5 rainy days a year and predicts rain incorrectly 36.5 dry days a year then only $\frac{5}{5+36.5} \approx 12.05\%$ of rain forecasts are in fact rainy days. Since the error rate is higher than the rainy rate, $100\%$ of rain forecasts being correct is impossible. –  Henry Sep 18 '12 at 21:08
Thank you for the clarification--I had misunderstood the analysis. –  whuber Sep 18 '12 at 21:12

This looks like a textbook Bayesian inference problem to me. It's even been tagged Bayesian.

[EDIT] However first we need to assume that weatherman's accuracy is independent of the weather in order to apply that hunch. It needs to be consistently 90% regardless of rain or no rain. If we can assume that then we can calculate as follows:

$$P(\text{rain | weatherman says rain})=\frac{P(\text{weatherman says rain |rain})P(\text{rain})}{P(\text{weatherman says rain})}$$

$$P(\text{weatherman says rain |rain})=\frac{90}{100}$$ from the text.$$P(\text{rain})=\frac{5}{365}$$ as per the text (on a local average anyway). $$P(\text{weatherman says rain})=P(\text{weatherman says rain, rain})+P(\text{weatherman says rain, no rain})=\frac{90}{100}\frac{5}{365}+\frac{10}{100}\frac{360}{365}$$You put it together and you get $$\frac{1}{9}\approx 11.11\%$$ so there is agreement. My guess is that this question is simply a textbook problem.

@Henry I'm too disreputable to comment directly in your answer space but I like your approach to finding an upper bound in a dependent case.

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(-1) There is nothing in the problem that says $P(\mbox{weatherman says rain}|\mbox{rain})=.9$. We only know the weatherman is correct 90% of the time, i.e. $P(\mbox{weather man says rain and it rains}) + P(\mbox{weatherman says no rain and it does not rain}) = .9$. –  guy Sep 18 '12 at 15:22
Also note that OP states in a comment that he came up with this problem on his own, so there is no sense in trying to get into the head of whoever wrote the flawed textbook solution. –  guy Sep 18 '12 at 15:37
Thanks for not actually voting me down. That said I see your point. Perhaps I can qualify this by stating that this is the solution if the weather man's accuracy is independent of rain (among other assumptions). –  MAB Sep 18 '12 at 16:18
Nevermind about the 'not downvoting it' part. I see that it was upvoted but you didn't like my independence assumptions being unstated (probably best to assume dependence is going to be my take-away). –  MAB Sep 19 '12 at 16:43