Probability of rain in desert when a reliable weatherperson predicts one

Question

Antonio has an exciting soccer game coming up.  In recent years, it has rained only 5 days each year in the city where they live.

Unfortunately, the weatherperson has predicted rain for that day. When it actually rains, she correctly forecasts rain 90% of the time. When it doesn't rain, she incorrectly forecasts rain only 10% of the time.

What is the probability that it will rain on the day of Antonio’s soccer game?

Answer 1

Let's say that R is the chance of it raining on a given day and FR is the chance of the weather forecaster calling for rain on a given day.

From Bayes' Theorem, P(R|FR) = [P(FR|R)*P(R)]/P(FR). In other words, the probability of it raining given a forecast for rain is [ (probability of the forecast for rain on days it actually rains) * (probability of it raining on a given day) ] / (the probability of the forecaster saying it will rain on a given day).

P(FR|R) = 0.9 (since it rains 90% of the time a forecaster predicts rain)

P(R) = 5/365 (days of the year it should rain)

We need to consider two cases for P(FR) and sum them together:
Case 1: The forecaster calls for rain on days it will rain
Case 2: The forecaster calls for rain on days it will not rain

For Case #1, the forecaster is right about 90% of the days it will rain. Thus,
P( Case 1 ) = 0.9 * 5/365.

For Case #2, the forecaster calls for rain on 10% of the days it won't actually rain. Thus,
P( Case 2 ) = 0.1 * 360/365

P(FR) = P( Case 1 ) + P( Case 2 )

Thus,

P(R|FR) = (0.9 * (5/365)) / ( (0.9 * (5/365)) + (0.1 * (360/365)) ) = 0.111...

Thus, if the forecaster predicts rain, it will rain ~11.1% of the time.