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Probability of Weather

  1. Aug 19, 2011 #1
    1. The problem statement, all variables and given/known data

    This isn't a homework question, and in fact I have not yet taken probability, so I know very little about it. With that in mind, any attempts to "lead me through" this problem will almost definitely fail, so if you could just tell me, that would be best.

    If the probability that it will rain any during the time 2-3 pm is 50/50 rain/sun, and the probability that it will rain any during the time from 3-4 pm is also 50/50, and the probability that it will rain any during the 22 remaining one hour blocks is 0, what is the probability that it will rain any over 24 hours?

    What if instead of 50/50, the rain chance for each of the two rainable hours is 30%?


    3. The attempt at a solution
    So, I was thinking, possible outcomes over favorable outcomes, right? And so it could either rain not at all, rain at 2-3 but not 3-4, rain at 3-4 but not 2-3, or rain both 2-3 and 3-4, all equally likely. This gives 3 favorables over 4, so 75 % chance of rain. But that seemed high. Thoughts?

    I wouldn't know how to calculate the 30% scenario.
     
  2. jcsd
  3. Aug 19, 2011 #2

    I like Serena

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    Hi Madame of Dark Magic! :smile:

    An alternative method of calculating the result, is to calculate the chance that it does not rain at all.
    Any thoughts on what that chance is?

    The same method would apply to the 30% scenario...
     
  4. Aug 19, 2011 #3
    The chance it won't rain at all in this first scenario is obviously 25%, according to the method I used. I truly don't know how to do the 30% case. I would really appreciate if someone could just tell me outright.
     
  5. Aug 19, 2011 #4

    I like Serena

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    Aww, all right.
    If the chance for it to rain in a specific hour is 30%, the chance for it not to rain is 70%.
    The chance not raining in either hour is 70% x 70% = 49%.
    So the chance for rain is 100% - 49% = 51%.

    More generally, if A and B are 2 independent events, we have:
    P(A and B) = P(A) x P(B)
    P(not A) = 1 - P(A)
     
  6. Aug 19, 2011 #5
    Cool. Thank you :)
     
  7. Aug 19, 2011 #6

    I like Serena

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    You do realize we've been stretching the rules a little?
    We're not supposed to give solutions outright in the homework/coursework section!
    But I guess in your case I could make an exception. :wink:
     
  8. Aug 19, 2011 #7
    as I understand it, you're not supposed to give answers outright to homework questions. This is not a homework question, but since it is simple, and undergrad material, I am supposed to post it here. It seems like it should be okay to answer this sort of question.

    Besides, having almost zero knowledge of probability, there was no way I could have known what the equation was.
     
  9. Aug 19, 2011 #8

    Ray Vickson

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    An important piece of missing information here is whether or not the two one-hour periods' rains are _independent_; that is, if it *is* raining during the 2-3 period, does that change the probability it will rain in the 3-4 period? If the weather conditions are *independent* between the two periods (meaning that the rain, or lack of it, from 2-3 has no effect on 3-4---a doubtful assumption) then your problem is like that of tossing a coin twice and counting the number of heads you get in two tosses. The probability of no rain in the day is the product of the probabilities of no rain each period, which is 0.5*0.5 = 0.25. Thus Pr{rain today} = 1 - Pr{no rain today} = 0.75. This type of analysis would still apply in case the chances are not 50-50; it is like tossing a biased coin twice. For example, if Pr{rain} = 0.30 in each period we have Pr{no rain} = 0.70 in each period. Then, Pr{no rain today} = 0.7*0.7 = 0.49, giving P{some rain today} = 1 - 0.49 = 051.

    At this point you might wonder: why did I multiply the probabilities of no rain, then subtract the result from 1? Why did I not multiply the probabilities of rain instead? Well, the latter would give the probability of rain in *both* periods, which is not what you asked.

    RGV
     
  10. Aug 20, 2011 #9
    I considered them independent. I wasn't trying to duplicate the weatherman or anything, just get some simple ideas about it. Thank you for your explanation, especially the part about using the no rain instead of the rain to do the calculation. :)
     
  11. Aug 21, 2011 #10
    To show how it is done the other way.

    Probability of rain in the 2 hour period
    P_rain(24)=P_rain(23) x P_rain(34) +P_rain(23) x P_Not_rain(34) +P-Not_rain(23)xP_rain(34)

    P_rain(24)= 0.3 x 0.3 + 0.3 x 0.7 + 0.7 x 0.3

    P_rain(24)= 0.09 + 0.21 + 0.21 = 0.51
     
  12. Aug 21, 2011 #11
    Neat-o.

    I'm taking probability 4442 starting next week. I sure hope I do okay, given my lack of basic prob-ability! Get it, pun? Gotta laugh...I'm nervous...
     
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