Probability Wind or Rain if it's Windy?

[Whalstib]

Homework Statement

30% chance of rain, 45% chance of windy, 62% chance rain OR wind.

Assuming it is raining what is the probability that it is windy?

Homework Equations

P(wind or rain)= P(wind)+P(rain) - P(wind & rain)

This does not work out the way I tried it...

The Attempt at a Solution

I used a Venn Diagram based on the original percentages..

The professor supplied one upon questioning that works out to:

17% chance rain, 32% chance wind, 13% chance wind or rain.

I think the way he worked this is:
P(wind OR rain)= P(wind)*P(rain) = .45*.30 = .135 ( he disregrads the .62 wind AND rain probability)
Now the
P(rain) = .30 - .135= .165,
P(wind)= .45 - .135= .315
P(wind OR rain) = .135

Now my Venn diagram would set up like:

16.5% chance of rain, 31.5% chance of wind, 13.5 % chance wind or rain.

The final answer would be [P(wind OR rain) = .135]/original P(rain)=.30) so
.135/.3 = .433.

This is the correct answer provided but the solution seems counter intuitive to me. Have I just stumbled upon the correct answer? Is this the simplest way to get the correct answer?

The professor supplied the P(wind or rain)= P(wind)+P(rain) - P(wind & rain) eqn. but that does not seem correct...

Final exam tomorrow and guaranteed there will be a 15 point problem similar to this! Any advice appreciated.

THANKS!

W

 

[Ray Vickson]

Homework Statement

30% chance of rain, 45% chance of windy, 62% chance rain OR wind.

Assuming it is raining what is the probability that it is windy?

Homework Equations

P(wind or rain)= P(wind)+P(rain) - P(wind & rain)

This does not work out the way I tried it...

The Attempt at a Solution

I used a Venn Diagram based on the original percentages..

The professor supplied one upon questioning that works out to:

17% chance rain, 32% chance wind, 13% chance wind or rain.

I think the way he worked this is:
P(wind OR rain)= P(wind)*P(rain) = .45*.30 = .135 ( he disregrads the .62 wind AND rain probability)
Now the
P(rain) = .30 - .135= .165,
P(wind)= .45 - .135= .315
P(wind OR rain) = .135

Now my Venn diagram would set up like:

16.5% chance of rain, 31.5% chance of wind, 13.5 % chance wind or rain.

The final answer would be [P(wind OR rain) = .135]/original P(rain)=.30) so
.135/.3 = .433.

This is the correct answer provided but the solution seems counter intuitive to me. Have I just stumbled upon the correct answer? Is this the simplest way to get the correct answer?

The professor supplied the P(wind or rain)= P(wind)+P(rain) - P(wind & rain) eqn. but that does not seem correct...

Final exam tomorrow and guaranteed there will be a 15 point problem similar to this! Any advice appreciated.

THANKS!

W

Using R for rain and W for wind, you are given P(R), P(W) and P(R or W). You want to get P(R & W). Surely the formula you wrote in allows you to do that!

 

[Whalstib]

Using R for rain and W for wind, you are given P(R), P(W) and P(R or W). You want to get P(R & W). Surely the formula you wrote in allows you to do that!

I'm not seeing it or doing it wrong:
P(w) = .45, P(r)= .30 P(r&w)=P(w)*P(r)= .45*.30=.135

so

P(w) = .45 -.135 =.32
P(r) = .30 - .135 = .165

and as presented
P(r OR w) = P(w) + P(r) - P(r&w) = .32 + .165 - .135 = .35 WRONG!

The correct answer is .433...where am I stumbling...?

Thanks,

W

 

[Ray Vickson]

I'm not seeing it or doing it wrong:
P(w) = .45, P(r)= .30 P(r&w)=P(w)*P(r)= .45*.30=.135

so

P(w) = .45 -.135 =.32
P(r) = .30 - .135 = .165

and as presented
P(r OR w) = P(w) + P(r) - P(r&w) = .32 + .165 - .135 = .35 WRONG!

The correct answer is .433...where am I stumbling...?

Thanks,

W

Why do you say P(R & W) = P(R)*P(W) = 0.135? Why do you then modify P(R) by subtracting 0.135 from the given value? None of what you are doing makes any sense.

 

[Whalstib]

Why do you say P(R & W) = P(R)*P(W) = 0.135? Why do you then modify P(R) by subtracting 0.135 from the given value? None of what you are doing makes any sense.

You noticed! Yes I am lost on this as the instructions do not match the way the prof figured it out. See my first post...

Unfortunatly this is an online course. I met with the prof once and got the explanation I provided. Working on my own I found the eqn that you agreed was the way to do it. I don't see how to do this and am quite lost. I have spent well over an hour and can only solve by the round about way I initially posted.

If you could walk me through it...

Thanks,

W

 

[Ray Vickson]

You noticed! Yes I am lost on this as the instructions do not match the way the prof figured it out. See my first post...

Unfortunatly this is an online course. I met with the prof once and got the explanation I provided. Working on my own I found the eqn that you agreed was the way to do it. I don't see how to do this and am quite lost. I have spent well over an hour and can only solve by the round about way I initially posted.

If you could walk me through it...

Thanks,

W

In your first post you wrote everything you need under heading 2: Relevant equations. Go back and read again what you wrote. (I really do want you to discover this for yourself, rather than showing it to you.)

 

[Whalstib]

In your first post you wrote everything you need. Go back and read again what you wrote. (I really do want you to discover this for yourself, rather than showing it to you.)

Thanks anyway Ray...

Anyone else care to point out where I am making my mistake? I'm not seeing it and trying to work through ~ 75 problems before the final.

This is the only one that I am clueless on.W

 

[Dick]

Thanks anyway Ray...

Anyone else care to point out where I am making my mistake? I'm not seeing it and trying to work through ~ 75 problems before the final.

This is the only one that I am clueless on.W

One mistake is that you can only take P(R & W)=P(R)*P(W) if the events R and W are independent. You can't assume that here. That's one reason it isn't working. Try to work it out without assuming that.

 

[Whalstib]

One mistake is that you can only take P(R & W)=P(R)*P(W) if the events R and W are independent. You can't assume that here. That's one reason it isn't working. Try to work it out without assuming that.

OK Based on my instructors

P(W or R) = P(W) + P(R) - P(W & R) = .30 + .45 - (.135) = .615

This is wrong, the answer is .433.

This is the only definition of the above eqn. I can find and it does not work.

Thanks,

W

 

[Whalstib]

One mistake is that you can only take P(R & W)=P(R)*P(W) if the events R and W are independent. You can't assume that here. That's one reason it isn't working. Try to work it out without assuming that.

So this means it can't be windy without rain or vice versa? Nonsense...they are independent in the real world. Who knows about the absurd world of statistics. Don't get me started on plots "skewed to the right" that are CLEARLY in ANY LANGUAGE SKEWED to the LEFT!

I think statistics is a joke. I got A's through Calc 2 with less effort and comprehended the math. Couldn't get in vector calc and have to suffer through this nonsense...sigh...and I graduate after this one...

W

 

[Ray Vickson]

So this means it can't be windy without rain or vice versa? Nonsense...they are independent in the real world. Who knows about the absurd world of statistics. Don't get me started on plots "skewed to the right" that are CLEARLY in ANY LANGUAGE SKEWED to the LEFT!

I think statistics is a joke. I got A's through Calc 2 with less effort and comprehended the math. Couldn't get in vector calc and have to suffer through this nonsense...sigh...and I graduate after this one...

W

You apparently totally misunderstand what is meant by "independence"---you have confused it with "mutual exclusiveness".

Of course you can have both R and W. Independence would mean that the chance of W on a rainy day is the same as the chance of W on a dry day, and similarly for R on a windy or still day. What we are saying is that:
(i) if R and W are independent, then P(R & W) = P(R)*P(W).
(i) if R and W are not independent, then P(R & W) ≠ P(R)*P(W).
Whether or not R and W are independent in this problem is not a matter of assumption; after computing P(A & B) we can test it.

In contrast, mutually exclusive events are just about as dependent as you can get: if A and B are mutually exclusive, they cannot occur together, so if A occurs then B does not, and vice-versa. They would have P(A & B) = 0, even though P(A) and P(B) are both nonzero.

 

[Whalstib]

You apparently totally misunderstand what is meant by "independence"---you have confused it with "mutual exclusiveness".

Of course you can have both R and W. Independence would mean that the chance of W on a rainy day is the same as the chance of W on a dry day, and similarly for R on a windy or still day. What we are saying is that:
(i) if R and W are independent, then P(R & W) = P(R)*P(W).
(i) if R and W are not independent, then P(R & W) ≠ P(R)*P(W).
Whether or not R and W are independent in this problem is not a matter of assumption; after computing P(A & B) we can test it.

In contrast, mutually exclusive events are just about as dependent as you can get: if A and B are mutually exclusive, they cannot occur together, so if A occurs then B does not, and vice-versa. They would have P(A & B) = 0, even though P(A) and P(B) are both nonzero.

OK..

SO can you solve it for me...?

W

 

[Whalstib]

You apparently totally misunderstand what is meant by "independence"---you have confused it with "mutual exclusiveness".

Of course you can have both R and W. Independence would mean that the chance of W on a rainy day is the same as the chance of W on a dry day, and similarly for R on a windy or still day. What we are saying is that:
(i) if R and W are independent, then P(R & W) = P(R)*P(W).
(i) if R and W are not independent, then P(R & W) ≠ P(R)*P(W).
Whether or not R and W are independent in this problem is not a matter of assumption; after computing P(A & B) we can test it.

In contrast, mutually exclusive events are just about as dependent as you can get: if A and B are mutually exclusive, they cannot occur together, so if A occurs then B does not, and vice-versa. They would have P(A & B) = 0, even though P(A) and P(B) are both nonzero.

I can sit down with my E&M experiments and equations and prove the magnetic flux mathematically and measure it to an uncanny degree of accuracy. In calc 2 I can integrate any odd shape (almost) and provide a highly accurate answer for its volume, measure and prove it to any one.

I come into stats and because one can’t prove wind and rain are independent you have to assume they are not to make the math work?

Stats is NOT mathematics but puzzles and games with arcane rules.

Sorry I am just beyond frustrated and can find no one that actually knows how to work this problem and is willing to share it with me. I have worked it out every way I have been taught and none provides the correct answer... I'm starting to think the answer provided .433 it in correct.

Could anyone please just work it out for me!

W

 

[Dick]

I can sit down with my E&M experiments and equations and prove the magnetic flux mathematically and measure it to an uncanny degree of accuracy. In calc 2 I can integrate any odd shape (almost) and provide a highly accurate answer for its volume, measure and prove it to any one.

I come into stats and because one can’t prove wind and rain are independent you have to assume they are not to make the math work?

Stats is NOT mathematics but puzzles and games with arcane rules.

Sorry I am just beyond frustrated and can find no one that actually knows how to work this problem and is willing to share it with me. I have worked it out every way I have been taught and none provides the correct answer... I'm starting to think the answer provided .433 it in correct.

Could anyone please just work it out for me!

W

I'm sure you are good at some other problems. Stats IS mathematics and there are no arcane rules here. If you would listen to the advice given and not keep asking someone to work it out for you might make a little more progress.

 

[Whalstib]

I'm sure you are good at some other problems. Stats IS mathematics and there are no arcane rules here. If you would listen to the advice given and not keep asking someone to work it out for you might make a little more progress.

Thanks...

 

[Whalstib]

I'm sure you are good at some other problems. Stats IS mathematics and there are no arcane rules here. If you would listen to the advice given and not keep asking someone to work it out for you might make a little more progress.

I am done with this problem...

Here's the real issue with it...the events are clearly not related. It is impossible to say 62% of wind or rain. The idea one would have to ignore reality, the reality that wind and rain are not dependent, and make a test based imaginary numbers to come up with an eqn that while it works is based on fallacy is hilarious to me!

Given the scientific and common sense FACT wind and rain are independent and you can not solve this problem accepting reality proves probability is just puzzles.

Another problem we have been given is with 80% certainty you know the answers to 6 problems set up a scenario where you answer 100% of the problems. This is impossible and absolute chance. The solution was some bizarre random number scheme that once again ignores reality which it seems is the only way statistic works...

W

 

[Dick]

Thanks...

At the risk of annoying you further, P(W or R) = P(W) + P(R) - P(W & R) is always true. You know P(W or R), P(W) and P(R). So you can find P(W & R). If your teacher's solution used P(W & R)=P(W)*P(R), that's simply wrong. Not arcane.

 

[Dick]

I am done with this problem...

Here's the real issue with it...the events are clearly not related. It is impossible to say 62% of wind or rain. The idea one would have to ignore reality, the reality that wind and rain are not dependent, and make a test based imaginary numbers to come up with an eqn that while it works is based on fallacy is hilarious to me!

Given the scientific and common sense FACT wind and rain are independent and you can not solve this problem accepting reality proves probability is just puzzles.

Another problem we have been given is with 80% certainty you know the answers to 6 problems set up a scenario where you answer 100% of the problems. This is impossible and absolute chance. The solution was some bizarre random number scheme that once again ignores reality which it seems is the only way statistic works...

W

Stats is about testing your ideas about what FACT is by dealing with the actual numbers. You could be wrong. Even though in an exercise the numbers might be made up instead of actually observed. Your idea of obvious reality might be wrong. That's what stats is supposed to test.

 

[willem2]

Given the scientific and common sense FACT wind and rain are independent and you can not solve this problem accepting reality proves probability is just puzzles.

I don't think you really mean what independent is. The fact that it's possible to have wind without rain, or rain without wind doesn't mean. It means that the probability that it rains if it is not windy, is equal to the probability that it rains if it is windy.

In real life, as opposed to statistics problesm, independent probabilities are rare.
Even if one event doesn't cause the other event, there are likely to be other events that cause both. I'm pretty sure that in most places, it's more likely to rain if it is windy. Low pressure areas and fronts often produce both wind and rain.
If you wanted to find out, you would have to look up weather data, and finally, you'd come up with statistics, such as in this problem.

 

[D H]

The correct answer is .433...where am I stumbling...?

You are stumbling because you only have one of the relevant equations at hand. The other equation is the definition of conditional probability, P(windy given that it is raining).

You are given the probabilities that it will be windy (45%), that it will rain (30%), and that it will be windy or rainy (62%). From these values you can determine the probability that it will be windy and rainy by the one equation that you did supply, and from that you can use the concept of conditional probability to answer the question at hand.

 

[Whalstib]

I don't think you really mean what independent is. The fact that it's possible to have wind without rain, or rain without wind doesn't mean. It means that the probability that it rains if it is not windy, is equal to the probability that it rains if it is windy.

In real life, as opposed to statistics problesm, independent probabilities are rare.
Even if one event doesn't cause the other event, there are likely to be other events that cause both. I'm pretty sure that in most places, it's more likely to rain if it is windy. Low pressure areas and fronts often produce both wind and rain.
If you wanted to find out, you would have to look up weather data, and finally, you'd come up with statistics, such as in this problem.

I'm sorry but this drives my point home dramatically.


The problem was simple; probability of wind if rain... If stats is mathematics/science it must be grounded in reality and if one must make a leap that wind and rain are dependent to solve the problem... well that takes it out of the realm of science and math and places it firmly in game and puzzle land.

It is taking qualitative measurements (define "windy") and making definite quantitative assumptions. I can easily contend the probability of it being windy while it rains as 100% based on MY definition of wind.


W

 

[Office_Shredder]

The problem was simple; probability of wind if rain... If stats is mathematics/science it must be grounded in reality and if one must make a leap that wind and rain are dependent to solve the problem... well that takes it out of the realm of science and math and places it firmly in game and puzzle land.

There's no leap of faith. There's no assumption at the start of the problem that wind and rain are correlated or independent or anything, there's just a couple of numbers. Along the way you prove that there is a correlation between wind and rain using well grounded mathematics. You're the one making some crazy leap of faith with "wind and rain must be independent!"

It is taking qualitative measurements (define "windy") and making definite quantitative assumptions. I can easily contend the probability of it being windy while it rains as 100% based on MY definition of wind.

That's great. Stats questions are like physics questions. They are going to describe systems that don't actually exist in the real world. Have you ever stood up and said "no particle can travel in exactly a parabolic motion! This is all games and puzzles" during a physics or calculus class? It's the same deal, they're describing a system that doesn't exist physically because it's easier to write problems if you don't have to take actual measurements. If the teacher said that he measured how many times it was windy ( and he said his definition: more than 3 mph of wind average over the day) and the number of times it was rainy (definition: at least one rain drop falls during the 24 hour period from midnight to midnight) and these are the numbers he got, would you accept the question as written?

The point of the question is: suppose someone took these measurements. What knowledge can you gain from it? And your answer is to spit on the measurements and declare them wrong.

 

[Whalstib]

There's no leap of faith. There's no assumption at the start of the problem that wind and rain are correlated or independent or anything, there's just a couple of numbers. Along the way you prove that there is a correlation between wind and rain using well grounded mathematics. You're the one making some crazy leap of faith with "wind and rain must be independent!"
That's great. Stats questions are like physics questions. They are going to describe systems that don't actually exist in the real world. Have you ever stood up and said "no particle can travel in exactly a parabolic motion! This is all games and puzzles" during a physics or calculus class? It's the same deal, they're describing a system that doesn't exist physically because it's easier to write problems if you don't have to take actual measurements. If the teacher said that he measured how many times it was windy ( and he said his definition: more than 3 mph of wind average over the day) and the number of times it was rainy (definition: at least one rain drop falls during the 24 hour period from midnight to midnight) and these are the numbers he got, would you accept the question as written?

The point of the question is: suppose someone took these measurements. What knowledge can you gain from it? And your answer is to spit on the measurements and declare them wrong.

A "crazy leap of faith with "wind and rain must be independent!"" You're killing me!

See you are arguing with semantics instead of proving a point with math...because clearly you can not. It is puzzles that change according to input. Not just the result changes but reality has to be either adjusted or defined and this is absurd.

Use numbers and get back to me...

Stats...sigh...

W

 

[D H]

I'm sorry but this drives my point home dramatically.

The point that your rants here and in your "statistics is not mathematics" thread drive home to me is that for some reason you don't want to learn the subject. Probability and statistics are a branch of mathematics, and they are very useful tools to many of the sciences.

 

[Whalstib]

The point that your rants here and in your "statistics is not mathematics" thread drive home to me is that for some reason you don't want to learn the subject. Probability and statistics are a branch of mathematics, and they are very useful tools to many of the sciences.

Not really...

I'm just waiting for the proof. So far it has been so much hot air and I contend it because it is puzzles and games. If it IS true math and science prove it with numbers not silly arguments.

Stats is a limited tool. There is not one problem I have had to solve this semester where the solution was self evident by looking at the data.

I love Physics Forum! Poseur scientists who want to argue yet rarely back up with facts or proofs..

Impress me, solve a problem...I have been wrong many times and will be the first to apologize and admit wrong. Solve a similar problem I put forth then we can talk..otherwise... who cares...

W

 

[Whalstib]

Not really...

I'm just waiting for the proof. So far it has been so much hot air and I contend it because it is puzzles and games. If it IS true math and science prove it with numbers not silly arguments.

Stats is a limited tool. There is not one problem I have had to solve this semester where the solution was self evident by looking at the data.

I love Physics Forum! Poseur scientists who want to argue yet rarely back up with facts or proofs..

Impress me, solve a problem...I have been wrong many times and will be the first to apologize and admit wrong. Solve a similar problem I put forth then we can talk..otherwise... who cares...

W

In fact I apologize right now...I'm being a jerk and I'm sorry. I should be using more careful words to express my misgivings on the subject. I am still unconvinced of much of it's usefulness as I have had to work through numerous difficult problems and get abstract results and on top of that have a professor who does not grade fairly. Multiple choice on math exams should be illegal! My calculator tends to use more decimal points and my prof will mark off for a P-value of .8884 when his TI says .8881...

Sorry...

BUt if you want to continue the discussion with some math I am more than happy to honestly engage you in a discussion.


W

 

[Office_Shredder]

Here's a fully rigorous mathematical statement. If every day I make a measurement to see if two things occur, and thing A happens 30% of the time, and thing B happens 45% of the time, and at least one of thing A or thing B happens 62% of the time, then thing A and thing B must both happen 13% of the time.

 

[D H]

I'm just waiting for the proof.

Proof of what?

Proof that probability and statistics are mathematics? That's the subject of Kolomogorov's axiomatization of probability theory. Simply put:

1. The probability of some event is a number between 0 and 1, inclusive.
2. The probability of the sample space as a whole is 1, exactly.
3. The probability of the union of a set of disjoint events is the sum of the probabilities of the individual events.

To understand Kolomogorov probability theory fully you'll need to understand measure theory. To understand that you'll need to know abstract algebra, real analysis, and complex analysis. It's a bit of a long road.

You want proof that probability and statistics are very important to science? Read any experimental physics journal. Read any scientific journal on diseases and cures for them. Read any scientific journal on engineering management. You're bound to run into lots of probability and statistics in those widely disparate journals. One example: The announcement last July of the discovery of the Higgs was heavily laden with probability and statistics. The LHC physicists waited to make that announcement official until they had the probability that there observations were a statistical fluke down to almost nothing (five sigma confidence level, IIRC).

If you don't like the wind and rain problem, change the labels. Call them events A (=rain) and B (=windy). Or ping pong balls with one hemisphere painted either red (=rain) or yellow, the other painted blue (=windy) or black. It doesn't matter what the labels are. The question is asking about a conditional probability, P(B|A). You did not reference the relevant equation for conditional probabilities. You need to do that to solve this problem.

 

[haruspex]

P(W or R) = P(W) + P(R) - P(W & R) = .30 + .45 - (.135) = .615

You obtained .135 by using the P(W & R) = P(W) * P(R) equation which you've already been told only applies if W and R are known to be independent. Please just accept that you must not make that assumption.

You are given the values of P(W), P(R) and P(W or R).
You have two equations:
P(A or B) = P(A) + P(B) - P(A & B)
P(A & B) = P(A|B) * P(B)
You are asked to calculate P(W|R)
Just rearrange the equations to match the problem structure and substitute the given values.