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Probability of rain watch

1. 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?

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2. (Original post by 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?

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There are only 4 possibilities, and raining on Sunday/Saturday are independent events from one another by the looks of it.

They are:
-Rain on Sat and rain on Sun
-Rain on Sat and no rain on Sun
-No rain on Sat and rain on Sun
-No rain on either day.

What's the probability that it rains on both? That would be rain on Sat x rain on Sun, so
What's the probability that is rains on Sat and not on Sun? Well that's
What's the probability it doesn't rain on Sat but rain on Sun? That's

Thus the total probability that it will rain on the weekend is 55% by adding up the first 3 possibilities.

The answer for the original one is 75% so you're right. The easiest way to calc it to realise each day has equal chance of raining, so over the two days, each possibility has an each chance of happening. 1/4=0.25 for any possibility, adding up the first 3 gives the answer right away.
3. You can calculate it using a tree diagram like so:

I converted the 50% to a fraction. When using probability, any "AND" statements you multiply, so I figured out the chance of it not raining on Saturday AND not raining on Sunday to be 1/4 (or 0.25). Probability always equals to 1, so I took 1/4 away from 1, getting 3/4 (or 0.75)

You could do it another way. Any OR statements are addition, so you multiply the AND statements.

Chances of raining on Saturday AND Sunday: 1/4
Chances of raining on Saturday AND dry Sunday: 1/4
Chances of being dry on Saturday AND rain of Sunday: 1/4

1/4 + 1/4 + 1/4 = 3/4 = 0.75
4. (Original post by Whispers)
You can calculate it using a tree diagram like so:

I converted the 50% to a fraction. When using probability, any "AND" statements you multiply, so I figured out the chance of it not raining on Saturday AND not raining on Sunday to be 1/4 (or 0.25). Probability always equals to 1, so I took 1/4 away from 1, getting 3/4 (or 0.75)
That's a wrong approach, but the simplicity of the problem gives the right answer nevertheless. A tree diagram implies the chance of rain on a Sunday depends on whether it rains the prev day or not, which is not true by the provided information.
5. (Original post by RDKGames)
That's a wrong approach, but the simplicity of the problem gives the right answer nevertheless. A tree diagram implies the chance of rain on a Sunday depends on whether it rains the prev day or not, which is not true by the provided information.
The tree diagrams implies four possibilities. The chance of rain on both days, the chance of rain on just Saturday, the chance of rain on just Sunday and the chance of no rain at all. You can put Sunday first and Saturday second and you would have the exact same tree diagram, and the exact same calculations and the exact same answers

This is one of the main methods they use to teach probability calculations at higher GCSE, I don't know about higher levels or if higher levels though
6. (Original post by RDKGames)
There are only 4 possibilities, and raining on Sunday/Saturday are independent events from one another by the looks of it.

They are:
-Rain on Sat and rain on Sun
-Rain on Sat and no rain on Sun
-No rain on Sat and rain on Sun
-No rain on either day.

What's the probability that it rains on both? That would be rain on Sat x rain on Sun, so
What's the probability that is rains on Sat and not on Sun? Well that's
What's the probability it doesn't rain on Sat but rain on Sun? That's

Thus the total probability that it will rain on the weekend is 55% by adding up the first 3 possibilities.

The answer for the original one is 75% so you're right. The easiest way to calc it to realise each day has equal chance of raining, so over the two days, each possibility has an each chance of happening. 1/4=0.25 for any possibility, adding up the first 3 gives the answer right away.

I think you used chance of rain on Sunday is 10% not 100%

But I think that's pretty much what I did, isn't it? Not sure.
7. (Original post by Hawksteinman)
I think you used chance of rain on Sunday is 10% not 100%

But I think that's pretty much what I did, isn't it? Not sure.
Ah yeah misread as 0.1. And yes it's pretty much what you've done.

The fastest way is indeed 1-(all unwanted possibilities) and since here it is only a single unwanted possibility, this is the quickest way. Otherwise, just add the first 3 up for it.
8. (Original post by RDKGames)
Ah yeah misread as 0.1. And yes it's pretty much what you've done.

The fastest way is indeed 1-(all unwanted possibilities) and since here it is only a single unwanted possibility, this is the quickest way. Otherwise, just add the first 3 up for it.
Ok, thanks!

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9. just google the weather forecast

no need for all this math
10. (Original post by Ikeo)

no need for all this math
But what's the probablity that the weather forceast is correct, and does it depend on whether it's forcasting rain or not. Oh, my head!
11. This is Britain, the probability of it raining is inversely proportional to how well waterproofed you are with coats, umbrellas, etc.

But most of the commentary here is correct - most probabilistic circumstances people encounter in everyday life are independent events, and most fail to account for this. This was also the crux of the Baby P conviction, which has since been proved to have been an incorrect application of the probability/statistics and was as a result overturned...some 20 years later as I recall.

It's the same reason why you aren't more likely to win the lottery if you played it previously. You are just more likely on the whole to win if you play it consistently, compared to someone who has only played once - but the chance of winning for every instance of playing remains the same (or well, similar, as the chance of winning differs every week depending on how many played as I understand).

As Terry Pratchett said - "million to one chances crop up nine times out of ten"

This is not a good reason to play the lottery, incidentally, as your chances of winning are still negligible and the net profit from winning is likely to be considerably lower than you would expect when you account for how much money you spent on lottery tickets.
12. Probability of rain in Manchester- 100%
Probability of rain in school holidays- 100%
Probability of sunshine a few days after uni or school term starts in September- 100% (except on Manchester)
13. Of course there is not the slightest chance of the events "rain on Saturday" and "rain on Sunday" being independent! So the chance of rain over the weekend cannot be calculated from the data supplied.

This is one of those examples that should never be used as an example!

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