# Does a weather forecast affect computation of probability of rain?

Suppose in recent years, it has only rained an average of 5 days a year. The weather forecast, which has an accuracy of 90%, predicts rain for tomorrow. What is the probability of rain tomorrow?

A: <1%
B: between 5 and 15%
C: between 15 and 80%
D: between 80 and 90%
E: > 90%

Thank you!

We can't do your homework for you; show us what you've done so far. The question is vague about what it means by the "accuracy" of the forecast, but the statement about average rain in recent years seems like a statement about prior probability, which should hint at what?..

You've been given two predictions for probability of rain. Can you find the mean and standard deviation of the data?

Unfortunately thats all the info that was supplied in the question. Mean and std dev cannot be found. 90% accuracy means the forecast is correct 90% of the time.

I initially thought P(rain) would be 5/365, since the forecast should have no effect on weather. However, it seems counter-intuitive knowing that the forecast has a 90% chance of being correct. I am confused because having a correct forecast is equivalent to having rain, so Im not sure whether there is a need to use both of the given probabilities.

So far, only probability laws of addition and multiplication as well as conditional probability have been covered, so the computation should not be too involved.

The mean and standard deviation can be found. You've got two data points: the historical probability of rain for any given day, and the forecast probability of rain.

I think I have got the answer using Bayes Theorem. P (rain | rain is forecast) was whats required

I think I have got the answer using Bayes Theorem. P (rain | rain is forecast) was whats required

That sounds reasonable, given that the average rain per year is essentially a prior probability.

I think I have got the answer using Bayes Theorem. P (rain | rain is forecast) was whats required

Bayes theorem applies when you've got two separate events whos probabilities are dependent. I don't think it applies here. P(rain|rain is forecast) can be stated as "probability of rain, given that rain is forecast." This is exactly what the question is asking you to find. In order to use Baye's theorem to find this, you need to know P(rain is forecast|rain), or "probability that rain was forecast, given that it rains." This isn't given in the problem.

I think this problem really is as simple as finding the mean and standard deviation.

You've been given two predictions for probability of rain. Can you find the mean and standard deviation of the data?

I'm really confused by this. Is there a hidden assumption that the weather forecast is independent of the 5 days statistic? Because it seems to me that the weatherman's 90% prediction would take all prior known data into account. So if someone asked me this question as given, I'd say the answer is 90% plus or minus the weatherman fudge factor.

Bayes theorem applies when you've got two separate events whos probabilities are dependent. I don't think it applies here. P(rain|rain is forecast) can be stated as "probability of rain, given that rain is forecast." This is exactly what the question is asking you to find. In order to use Baye's theorem to find this, you need to know P(rain is forecast|rain), or "probability that rain was forecast, given that it rains." This isn't given in the problem.

I think this problem really is as simple as finding the mean and standard deviation.

The question is isomorphic to the "canonical" question requiring Bayes' theorem: Only 5 out of 365 people have breast cancer. Given that breast cancer screening is 90% accurate, what is the probability that someone who tests positive actually has breast cancer?

I think this problem really is as simple as finding the mean and standard deviation.

Of what? The probability of rain? You still need the prior probability of rain and the accuracy of the forecast. You could take the classic Bayesian approach and treat the parameter of interest as a random variable, but then there's a fair bit more to in than simply "calculating the mean and SD".