Results 1 to 14 of 14

Math Help - Probability Problem

  1. #1
    Junior Member mathhomework's Avatar
    Joined
    Jan 2009
    Posts
    37

    Probability Problem

    In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by mathhomework View Post
    In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
    Form a list of 7 empty spaces:
    _ _ _ _ _ _ _

    If it rains on Sunday, Monday, Tuesday, Wednesday, and Friday you write:
    X X X X X _ _

    There are {7\choose 5} = {7\choose 2} = \frac{7\cdot 6}{2} = 21 different ways to put five X's.

    The probability that it does not rain on the last two days (I assume that is what you mean by 'weekend') is getting the first five X's and the last two empty. There is only one way to do that. Therefore, the probability that it would not rain is \tfrac{1}{21}. Thus, the probability that it would rain is \tfrac{20}{21}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mathhomework View Post
    In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
    My take on this is the following:

    Let X be the random variable number of days on the weekend that it rains.

    X ~ Binomial(n = 2, p = 5/7).

    Calculate Pr(X = 1).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    My take on this is the following:

    Let X be the random variable number of days on the weekend that it rains.

    X ~ Binomial(n = 2, p = 5/7).

    Calculate Pr(X = 1).
    My answer has assumed independence. If that's not the case, then more information is required to answer the question.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    49
    Quote Originally Posted by ThePerfectHacker View Post
    Form a list of 7 empty spaces:
    _ _ _ _ _ _ _

    If it rains on Sunday, Monday, Tuesday, Wednesday, and Friday you write:
    X X X X X _ _

    There are {7\choose 5} = {7\choose 2} = \frac{7\cdot 6}{2} = 21 different ways to put five X's.

    The probability that it does not rain on the last two days (I assume that is what you mean by 'weekend') is getting the first five X's and the last two empty. There is only one way to do that. Therefore, the probability that it would not rain is \tfrac{1}{21}. Thus, the probability that it would rain is \tfrac{20}{21}.
    You are correct. However, that doesn't answer the question. The question is what is the probability that it will rain exactly ONE day on the weekend. You have determined the probability that it will rain on the weekend period. There are of course, three ways for that to happen: it can rain on Saturday, on Sunday, or on both. The number of total possible outcomes is simply 7!/5!/2! = 21 Fixing a rain day on Saturday and dry day on Sunday, we have 5!/4! = 5 outcomes. Doing the same for a rain day on Sunday and dry day on Sat. we have 5!/4!=5 outcomes. The probability is 10/21
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Bilbo Baggins View Post
    You are correct. However, that doesn't answer the question. The question is what is the probability that it will rain exactly ONE day on the weekend. You have determined the probability that it will rain on the weekend period. There are of course, three ways for that to happen: it can rain on Saturday, on Sunday, or on both. The number of total possible outcomes is simply 7!/5!/2! = 21 Fixing a rain day on Saturday and dry day on Sunday, we have 5!/4! = 5 outcomes. Doing the same for a rain day on Sunday and dry day on Sat. we have 5!/4!=5 outcomes. The probability is 10/21
    I would dispute this answer. The probability of rain on any given day is 5/7. It's not a matter of listing outcomes .....
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2008
    Posts
    52
    I think Bilbo is right. If it rains on saturday you have 4 pegs, let's say, to put on 5 different days, mon, tues, weds, thurs, friday.

    there are 5 ways to do it as you can only miss one day out each time, and there are 5 days.

    Then you double this as you could also put the first peg on sunday.

    so you have the total number of ways you can get the outcome, divide it by the total number of outcomes and you have 10/21
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Bruce View Post
    I think Bilbo is right. If it rains on saturday you have 4 pegs, let's say, to put on 5 different days, mon, tues, weds, thurs, friday.

    there are 5 ways to do it as you can only miss one day out each time, and there are 5 days.

    Then you double this as you could also put the first peg on sunday.

    so you have the total number of ways you can get the outcome, divide it by the total number of outcomes and you have 10/21
    5 days out of 7 is only an average, it does not mean that it does rain 5 days a week. It only means that the probability of rain on a given day is 5/7. There's a big difference.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2009
    Posts
    49
    Quote Originally Posted by mr fantastic View Post
    5 days out of 7 is only an average, it does not mean that it does rain 5 days a week. It only means that the probability of rain on a given day is 5/7. There's a big difference.
    True. However, I think that you are looking into the problem too deeply. A problem of this type usually doesn't require that type of scrutiny. I think that saying "rains an average of 5 times a week" is just code for "rains 5 times a week." Perhaps, because this is a statistics problem they didn't want to sound too unequivocal and added "on average"
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member mathhomework's Avatar
    Joined
    Jan 2009
    Posts
    37

    Thank You

    Thanks everybody for helping me!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Bilbo Baggins View Post
    True. However, I think that you are looking into the problem too deeply. A problem of this type usually doesn't require that type of scrutiny. I think that saying "rains an average of 5 times a week" is just code for "rains 5 times a week." Perhaps, because this is a statistics problem they didn't want to sound too unequivocal and added "on average"
    Then I suppose we will agree to disagree until the answer is given by the OP.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Dec 2008
    Posts
    52
    My friend at school gave his take on the question:

    there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49

    the same for rain on sunday and none on saturday, so you end up with 20/49

    this seems a plausible answer.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Bruce View Post
    My friend at school gave his take on the question:
    there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49 the same for rain on sunday and none on saturday, so you end up with 20/49. This seems a plausible answer.
    I absolutely agree with that interpretation of this problem.
    And I think that \frac{10}{49} is correct.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Bruce View Post
    My friend at school gave his take on the question:

    there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49

    the same for rain on sunday and none on saturday, so you end up with 20/49

    this seems a plausible answer.
    Which exactly follows from what I posted way back in post #3 since \Pr(X = 1) = ^2C_1 \left(\frac{5}{7}\right)^1 \left(\frac{2}{7}\right)^1 = \frac{20}{49}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability problem
    Posted in the Statistics Forum
    Replies: 4
    Last Post: January 30th 2011, 07:16 PM
  2. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  3. Probability problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 30th 2009, 02:40 PM
  4. Replies: 0
    Last Post: October 8th 2009, 08:45 AM
  5. probability problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 16th 2008, 12:45 PM

Search Tags


/mathhelpforum @mathhelpforum