Probability of Rain During a Collecting Trip

[DPS Ammonite]

I have a mathematics question that is tangentially related to fossil collecting. How do you figure out the total chance of rain during a collecting trip using the weather service rain predictions (should I bring my umbrella)? Can you give a general formula? Let us say that you are going on a three day trip with the following precipitation chances of measurable rain (chance of rain for any geographical point in forecast area) for each day: Friday 10%; Saturday 20% and Sunday 30%. What is the combined probability, Friday through Sunday, that it will rain at least once? Is rain likely? Assume that chance of rain for any period is independent of each other. 

 

I know that the total chance is at least as high as the highest chance for one day: 30%. It is lower than adding 10, 20 and 30 equals 60%. 

 

What is the % answer and the general formula? 

 

Thanks,

John

 

EDIT.     I found a website post below with a similar question: There is a 20% chance of rain each day for 5 days; the chance of rain during the whole period is 67%. 

 

I did solve the problem correctly before I found website below, I think.

 

Solve the problem by using the chance that it is not going to rain each day in decimal form: .9 x .8 x .7 = .504 or 50.4% chance that it is not going to rain. Therefore there is a 49.6 % chance of rain. Bring an umbrella.

 

https://www.theweatherprediction.com/habyhints/266/

 

 

PRECIPITATION PROBABILITY BRAIN TEASER

 

METEOROLOGIST JEFF HABY

A broadcast meteorologist gives the following forecast:

Monday: 20% chance of rain
Tuesday: 20% chance of rain
Wednesday: 20% chance of rain
Thursday: 20% chance of rain
Friday: 20% chance of rain

A viewer is having a week long outdoor event that lasts from Monday through Friday. Monday morning the viewer asks the broadcast meteorologist what the chance for rain is for the entire week as a whole. In other words the viewer wants to know what the chance is it will rain on either Monday, Tuesday, Wednesday, Thursday, or Friday.

What is the answer? Assume the probability of precipitation (POP) is independent for each day and the forecasted POP does not vary with time.

SOLUTION: This situation represents the probability of rain within a 5 day period given at the beginning of the week and assumes each day is an independent probability. Two of the answer choices can be eliminated through applicable logic. It is known the probability of rain during the week is greater than 20% since each day has at least a 20% chance. It is also known the probability can not be 100% because the possibility is clearly evident that it might not rain at all during the week.

A probability for this case is solved by multiplying the probability it will not rain each day and subtract this from 100%. The left over value is the chance that is will rain during the week. The chance of no rain each day is 80%. Thus the chance for no rain each day put together is: 0.8 * 0.8 * 0.8 * 0.8 * 0.8 = 0.33 or 33%. Since the chance for no rain is 33%, the chance for rain is 100% - 33% = 67%.

(1 - 0.8^5) = 1 - 0.33 = 0.67 * 100% = 67%

Thus, there is a 67% chance (or a 2 in 3 chance) that the viewer will have rain sometime during the week.

[Darktooth]

I always bring a raincoat just in case

Just assume it will.

[Kane]

In terms of probability, it may not be plausible to assume these are independent events as weather is strongly correlated. But to go along with this example... It is a good idea to think of this question in terms of what else happens (i.e., the Event E where it does not rain).

 

D= day it rains, be it days 1, 2, 3

~D = day it does not rain, be it days 1, 2, 3

 

P(rain) = P(D1,D2,D3) + P(D1,D2~D3 v ~D1,D2,D3) + P(D1,~D2,D3 v ~D1,D2,~D3 v ~D1,~D2,D3) v P(rain) = 1 - P(no rain)

 

[Here it is simply the either/or of whether it rains on any particular day, which is like the fair coin of 50/50, netting us a number of possibilities of rain and adding in complementarity. Keep in mind that, in probability, P1 is probability that absolutely will happen, and P0 is probability zero that an event happens. This becomes complicated when we try to compute probabilities more than two]

 

= 1 - P(~D1,~D2,~D3)

= 1 - P(~D1) x P(~D2) x P(~D3)

= 1 - 0.1 x 0.2 x 0.3 [these are your stated percentages as above expressed decimally; as a formula we would treat these as a, b, c...n]

= 1 - 0.006

= 0.994

 

So, the probability it will not rain is quite high at P(rain) = 0.994. This may seem counterintuitive, but again it is important to think of the percentage probability of it not raining in your example (90, 80, and 70%, respectively). 

 

It is simpler with a two day example. Assume 70 and 90% chance of rain on Friday (F) and Saturday (S). What are the aggregate chances of rain over that two day period?

 

0.3 x 0.9 [chance it won't rain on Friday = 30% and 90% chance it will on Saturday]

0.7 x 0.1 [chance it rains on Friday and chance it doesn't on Saturday]

0.7 x 0.9 [chance it rains on both days]

 

= 1 - (0.3 x 0.9) + (0.7 x 0.1) + (0.7 x 0.9)

= 1 - (0.27)+(0.07)+(0.63)

 

97% chance it will rain at least one of the two days.

 

 

 

[Kane]

2 hours ago, DPS Ammonite said:

A probability for this case is solved by multiplying the probability it will not rain each day and subtract this from 100%. The left over value is the chance that is will rain during the week. The chance of no rain each day is 80%. Thus the chance for no rain each day put together is: 0.8 * 0.8 * 0.8 * 0.8 * 0.8 = 0.33 or 33%. Since the chance for no rain is 33%, the chance for rain is 100% - 33% = 67%.

(1 - 0.8^5) = 1 - 0.33 = 0.67 * 100% = 67%

Thus, there is a 67% chance (or a 2 in 3 chance) that the viewer will have rain sometime during the week.

You posted while I was posting! Yes, you have to subtract from P(1). When taking a longer time series, the probability of an Event E approaches closer to 1. But not on any individual instance. For example, the chance of being in a plane crash on any particular trip is extremely low, but over time the probability increases with multiple trips even if each trip's chances of a crash remain the same. Put another way, the probability of rolling a 44 on a 100-sided die is 1/100, but keep rolling the dice and eventually 44 will turn up even though the individual chances remain 1/100.  

[minnbuckeye]

Statistics . Even though mathematically inclined (love calculus), I will never enjoy the subject. So like @Darktooth says, I will just take the raincoat!

 

Mike

[RJB]

Im with DarkTooth, I always bring a rain coat and 2 umbrella's.   One umbrella can break?

 

RB

[DPS Ammonite]

38 minutes ago, Kane said:

 

= 1 - P(~D1,~D2,~D3)

= 1 - P(~D1) x P(~D2) x P(~D3)

= 1 - 0.1 x 0.2 x 0.3 [these are your stated percentages as above expressed decimally; as a formula we would treat these as a, b, c...n]

= 1 - 0.006

= 0.994

 

So, the probability it will not rain is quite high at P(rain) = 0.994. This may seem counterintuitive, but again it is important to think of the percentage probability of it not raining in your example (90, 80, and 70%, respectively). 

Is your conclusion to my problem that there is a 99.4% chance of no rain over the three day period? If so, that defies logic. The probability of rain cannot be lower that the highest probability for any one period: 30%.

 

Did I solve my problem correctly: 49.6% chance of rain?

[Kane]

21 minutes ago, DPS Ammonite said:

Is your conclusion to my problem that there is a 99.4% chance of no rain over the three day period? If so, that defies logic. The probability of rain cannot be lower that the highest probability for any one period: 30%.

Whoops! I didn't add all of the variables after the disjunctions.  

Using that would net us the discrete instances, but the aggregate would indeed be 50.4% chance of no rain. [Pours another cup of coffee ]

[daves64]

Just bring scuba gear to be on the safe side. 

[Uncle Siphuncle]

One source of error in this method is confidence level of each daily prediction the farther into the future we look.  Seems that daily predictions aren’t worth a hoot more than 2 days out.  So you do your math 5 days out, then find that the weatherman starts flip flopping as the latter days approach.  No matter though.  Even though Monday’s forecast may change on a dime, some are certain the world will end in 12 years.  So just go out and have fun now.

[Fossildude19]

Wait a second. I was told there would be no math!   

 

I want to be a weatherman for my next job. 

The only one where you can be wrong 80% of the time, and still keep your job.      

[JohnJ]

The answer is to monitor the weather when there is a chance of rain. Even if there is a small chance each day...if you are in the spot it develops, then you will be 100% wet.  Calculating the total percentage across the 3 days, in the first post, is practically of little use in a real world scenario.  ☔

[FossilDAWG]

I quite like being able to pull up the weather radar on my phone and being able to see if, for example, that thunderstorm in the distance is moving towards me or away.  Short term I know but useful.

 

If the chance of rain is 10% and I forget my rain gear, the odds are 100% it will rain on me.

 

Don

[daves64]

24 minutes ago, FossilDAWG said:

If the chance of rain is 10% and I forget my rain gear, the odds are 100% it will rain on me.

Don't forget the rain gear and DON'T wax your car before going, should be drought conditions then.

[FossilDAWG]

1 hour ago, daves64 said:

... DON'T wax your car before going, should be drought conditions then.

What is this "wax your car" of which you speak?  I live on a farm on a dirt road.  My truck is lucky if I take a hose to it once or twice a year.

 

Also we are in a semi-drought here, it's rained maybe twice since June.  I've left the truck windows rolled down, hid my raincoat and umbrellas in the basement, left my most expensive tools out on the back deck, anything I can think of to make it rain.  Nothing is working.

 

Don

[daves64]

18 minutes ago, FossilDAWG said:

What is this "wax your car" of which you speak?  I live on a farm on a dirt road.  My truck is lucky if I take a hose to it once or twice a year.

 

Also we are in a semi-drought here, it's rained maybe twice since June.  I've left the truck windows rolled down, hid my raincoat and umbrellas in the basement, left my most expensive tools out on the back deck, anything I can think of to make it rain.  Nothing is working.

 

Don

Wax your truck or plan an outdoor event. Both are usually good way's to cause rain. Or wax your truck during an outdoor event. Sleep outside without a tent. Have the in-laws over for a visit. 

[DPS Ammonite]

4 hours ago, JohnJ said:

The answer is to monitor the weather when there is a chance of rain. Even if there is a small chance each day...if you are in the spot it develops, then you will be 100% wet.  Calculating the total percentage across the 3 days, in the first post, is practically of little use in a real world scenario.  ☔

I realize that weather forecast can change over a period of three days. When I first thought of the question, I was thinking if applying it to a period of a day or a day and a half. 

While most forecasts give the probabilities over a period of 12 hours, some give probabilities of rain over intervals of 1 hour for

1 day. Let’s say that there is a probability of rain of 3% per hour for 1 day. Should we concerned about getting wet over a 24 hour period?

 

 

 

 

Yes. (.97) to the 24th power = 48% chance of rain.

[JohnJ]

Or...umm, no.  I've been in plenty of scenarios where chances of rain were very low and endured a down pour; and in other cases where the chances were high, I listened to thunder and smelled the rain, but stayed dry.  Math has limitations when applied to weather. 

[FossilDAWG]

I was confused and curious so I looked up what "probability of precipitation" actually means.  It is the estimated probability that precipitation (POP) (at least 0.01 inch) will occur anywhere in the area covered by the forecast, multiplied by an estimate of the % of the area (within the area covered by the forecast) that will experience rain.  For example, say the meteorologist calculates that the chance of rain within her forecast area is 50%, and 80% of the area is likely to receive some of that rain.  The POP within the forecast area is 50% x 80% (0.5 x 0.8) = 40%.  That more-or-less corresponds to the chance that you will see rain if you stay at one spot all day.  Still, it seems to me that there is a large "fudge factor".  Anyway a 40% POP means a 60% chance that it won't rain, and being a "glass half full" sort of a person I prefer to look at it that way.  Besides, in many cases when the ground is wet it makes fossils easier to see.  As long as the forecast is not for a life-threatening deluge I'm not likely to be deterred.

 

Don

[Kane]

I asked my mathematics mentor who speaks of the difference between a meaningful definition of probability, and the fuzziness of applying it to weather (2nd para): 

 

"It's kind of a weird thing, given that as I understand it the probability of rain means the probability that rain falls on some location in the area. It is not about your specific location. So you first of all have to know what particular area they are talking about. Then it is geographical sort of thing. But take a look at the problem of rolling two fair dice. Then there are 36 possible events, each equally likely: {(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1),(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2),(2,6),(6,2),(3,3),(3,4),(4,3),(3,5),(5,3),(3,6),(6,3),(4,4),(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)}. So the probability of getting at least one six, for example, is obtained by counting up all the occurrences of 6: sum of the elements in the set {(i,j):i=6 or j=6) which is the union of the singleton sets {(i,j)|i=6 or j=6} = number of elements in {(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5),(6,6)} = 11 and dividing by 36. Or one could simply note that if 6 occurs with probability 2/36 for each of the numbers 1 through 5 and twice with 6, so it is 2*5/36 + 1/36. (One can change the problem by painting the dice different colors and worrying about the order, but that has noting to do with the numbers: there is one way to get (1,1) = 2, while there are ways to get 3, namely 1 first or 2 first, but if one can distinguish dice and asks about getting 2 with blue first and red second, that is a different question than we are concerned with and adds 6 more events, (1 b,i r) versus (i r,i b  ) for i =1 to 6.) The probability of getting at exactly one 6 is 10/36. The probability of getting a 6 on the first toss is 6/36 or 1/6. This has to do with disjoint events so they can be counted once each. Questions that mix stuff up like the probability of getting sum of 4 with a product of 3  from the two sides can be thought of as being {(1,3),(3,1),(2,2)} = {(i,j)|i+j = 2} minus the set {(i,j)|ixj=4} = {(2,2),(4,1),(1,4)} which is the same as the intersection of {(1,3),(3,1),(2,2) with the complement of {(2,2),(4,1),(1,4)}.

 

"That is the standard of discrete probability which is the only place where there is some sort of meaningful physical definition of probability. How one applies this to weather is a problem for me. A forecast of 10% chance of rain as I understand it means that anyone in the area the forecast covers has a .1 probability of experiencing rain (which I guess could mean 2 drops). But does that mean that if it's .1 one day and .4 the next day, YOU have a chance of .5 as the sum of two independent events (ie disjoint events)? No, not as I see it, as it doesn't apply to a specific location. I have a tough time wrapping my head around the events, actually. The event (the house at n xyz street getting rain) is not what the forecast is actually about. It is about (some location in the region in which n xyz street is located getting rain) is different than that. So it seems it would be .5 that some location gets rain, but some location does not necessarily refer to you at all and it also is not very well defined because there is no notion regarding just how big the location to get rain might be. So I am not sure about just what that assignment of probability means: I would have to get a very precise definition in terms of region to which it applies and how small the location might be etc."

[Uncle Siphuncle]

I've been out before when a 10% chance of .1 inch accumulation almost left me sprawled for eternity in the alluvial fan.  No fun when the leading wall of trash and sticks and mud is coming your way.  Know the drainage patterns where you collect, and hedge your bets accordingly.  I don't take too many chances in certain stream channels these days.  (But I made a couple killer finds that day ahead of the storm)

[Randyw]

9 hours ago, Fossildude19 said:

Wait a second. I was told there would be no math! 

I agree! After reading this my brain hurts! I hate math!

[Randyw]

1 hour ago, Uncle Siphuncle said:

I've been out before when a 10% chance of .1 inch accumulation almost left me sprawled for eternity in the alluvial fan.  No fun when the leading wall of trash and sticks and mud is coming your way.  

Given that flood conditions and rapid burial in mud increases the chance of fossilization I wonder what would have been the chances of someone in the future finding a fossilized uncle would have been if you had been slower? (And yes my brain still hurts from all the previous math) 

[Adam86cucv]

If professor Murphy has anything to do with it you'll have rain when you aren't prepared for it.   I can personally guarantee you that on the days you will be out there will be a 100% chance of weather or your money back. 

[FossilNerd]

11 hours ago, Fossildude19 said:

Wait a second. I was told there would be no math!   

 

I want to be a weatherman for my next job. 

The only one where you can be wrong 80% of the time, and still keep your job.      

So true! I have to constantly keep an eye on the weather forecast for my job and it’s wrong most of the time. Around here I don’t even start worrying about it until the chance gets 50% or above. There is a local saying that goes “If you don’t like the weather in Kentucky just wait 5 minutes.”