# B Probability of rain?

1. Jul 24, 2017

### Hawksteinman

I was reading a book about innumeracy and one of the chapters was on probability. This weather woman said 'there is a 50% chance of rain on Saturday, and a 50% chance of rain on Sunday, so the chance of rain this weekend is 100%'

Obviously she was wrong, but it got me thinking how would one calculate the probability of rain that weekend?

I decided to make it simpler by saying P(rain on Saturday) = 0.5 and P(rain on Sunday) = 1.0

This obviously means that P(rain this weekend) = 1.0

I then used trial and error to calculate the chance of rain that weekend

I started with P(rain on Saturday) x P(rain on Sunday) but that gives 0.5

Then I tried P(no rain this weekend) = P(no rain on Saturday) x P(no rain on Sunday) and this gives 0.0

Therefore P(rain this weekend) = 1 - P(no rain this weekend) = 1.0

Using this method, P(rain this weekend) = 0.75 or 75% for the original statement.

Is this the best way to calculate the chance of rain?

2. Jul 24, 2017

### Stephen Tashi

For events "A" ,"B"

probability of "A or B" = probability of "A" + probability of "B" - probability of "A and B"

For independent events , probability of "A and B" = (probability of "A")(probability of "B").

However, if it rains on Saturday, that may make it statistically more likely that it will also rain on Sunday, so those two events may not be independent.

In general, probability of "A and B" = ( probability of "B given A")(probability of "A").

Rain forecasts do state probabilities for rain, but "the probability of rain" is an ill-defined concept until we define the "probability space" we are using.

3. Jul 24, 2017

### Hawksteinman

Assuming that they are independent, am I right to do what I did? :)

4. Jul 24, 2017

### mjc123

How did you get 0.75? Averaging the two answers you got, one right, one wrong? That's not a good method!

5. Jul 24, 2017

### Hawksteinman

P(no rain on Saturday) x P(no rain on Sunday) = P(no rain this weekend) = 0.5 x 0.5 = 0.25

P(rain this weekend) = 1 - P(no rain this weekend) = 1 - 0.25 = 0.75

6. Jul 24, 2017

### scottdave

From what I have read and witnessed the TV weather forecaster state, this is how they define percentages. A rain percentage of 20% means that they expect that 20% of the viewing area will have rain. So if you are watching it, then you have a 20% chance of living in the area that will get rain. Sometimes they show predictive maps, which give you a better idea if your area falls in the 20% though.

But if the only information you have is 50% chance of rain on Saturday, then I would go with that - a straight probability of 1/2.

Use either method that @Stephen Tashi gave and you should come up with the same percentage for "the weekend".
(rain Saturday) Or (rain Sunday)
or you could do this: 1 - ((not rain Saturday) AND (not rain Sunday))

7. Jul 24, 2017

### Staff: Mentor

This is a correct calculation for independent events.

As @Stephen Tashi mentioned the key question for actually applying this would be if the percentages are independent or not. If there were definitely a very brief storm coming, but only the time of arrival was uncertain then it could be as high as 100% (perfectly anti correlated). If there was a possible storm coming but it would last both days if it came then it could be as low as 50% (perfectly correlated).