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Probability Wind or Rain if it's Windy?

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    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?



    2. Relevant equations

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

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


    3. 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
     
  2. jcsd
  3. May 12, 2013 #2

    Ray Vickson

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    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!
     
  4. May 12, 2013 #3


    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
     
  5. May 12, 2013 #4

    Ray Vickson

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    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.
     
  6. May 12, 2013 #5
    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
     
  7. May 12, 2013 #6

    Ray Vickson

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    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.)
     
    Last edited: May 12, 2013
  8. May 12, 2013 #7

    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
     
  9. May 12, 2013 #8

    Dick

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    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.
     
    Last edited: May 12, 2013
  10. May 12, 2013 #9
    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
     
  11. May 12, 2013 #10
    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
     
  12. May 12, 2013 #11

    Ray Vickson

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    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.
     
  13. May 12, 2013 #12
    OK..

    SO can you solve it for me....?

    W
     
  14. May 12, 2013 #13
    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 any one please just work it out for me!

    W
     
  15. May 12, 2013 #14

    Dick

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    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.
     
  16. May 12, 2013 #15
    Thanks...
     
    Last edited: May 12, 2013
  17. May 12, 2013 #16
    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
     
  18. May 12, 2013 #17

    Dick

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    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.
     
  19. May 12, 2013 #18

    Dick

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    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.
     
  20. May 13, 2013 #19
    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.
     
  21. May 13, 2013 #20

    D H

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    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.
     
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