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Basic probabilities

  1. Jun 17, 2017 #1
    1. The problem statement, all variables and given/known data
    the administration council of one industry is composed by 7 people 4 mens and 3 womens.
    an election its hapening to vote chairman and secretary, if all members of the council are candidates obtain the follow
    a) just mens take positions
    b) a chairwoman elected
    C) at least one woman is elected



    2. Relevant equations


    3. The attempt at a solution
    so far my work its the next
    i ve determined that there is 42 forms of people take positions, those forms are under 21 pairs of chairman and secretary, this because for the a and c, order is meaningless, just b requieres order,
    well my calculations end here: there are 3 pairs made of entirely womans and 6 pairs of entirely mens,
    so i supose there is 12 pairs of mixed positions.
    acording to my calculations
    a) 6/21= 28% of just mens being elected
    c) 3+12 the pairs of 3 womens and the mixed ones so 15/21=71% of at least a woman being elected
    i dont know if i am right, thats the reason i am asking for help, also i dont know how to calculate de b)
     
  2. jcsd
  3. Jun 17, 2017 #2
    Hi Mrenko:

    I think something is missing from the problem statement. What is the probability distribution of a person's vote. For example, is it equally likely that each person will vote for each of the seven candidates?

    Regards,
    Buzz
     
  4. Jun 17, 2017 #3
    Yes it is, is the same probabilities of vote, and everyone it's equally to everyone
     
  5. Jun 17, 2017 #4
    Hi Mrenko:

    If I understood correctly what you did, you assumed each pair of winners is equally likely to occur. Since a given pair is the result of two separate elections, are you sure this assumption is correct?

    Regarding (b), this is a single election case, since only the chair election is considered. If you keep that in mind, then this case is simpler than (a) and (c).

    Hope this helps.

    Regards,
    Buzz
     
  6. Jun 17, 2017 #5

    Ray Vickson

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    The events in (a) and (c) are complementary, so if you know the probability in one case you can get it right away in the other case as well.
     
  7. Jun 18, 2017 #6
    hi, i did this in pairs because i needed to know how much pairs of mens are there, and how much pairs of woman as well,
    so i know that there is a diference between the two positions to be voted, so should i just do a permutation of 7 to 1 to found the b?
    i dont get very well.
    for the second response, i have issues whit the problem being divided in two positions to be voted and two genders, i dont get it as a whole, sorry need help whit that
     
  8. Jun 18, 2017 #7

    Ray Vickson

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    In part (a) you should use permutations rather than combinations, because the cases where Mr. A is president and Mr. B is secretary is different from the case where Mr. B is president and Mr. A is secretary. In part (b) there is no need to look at pairs at all:---think about why!
     
  9. Jun 18, 2017 #8
    But is just asking the probabilities of a man being elected, don't matter if it's mr j,b,g,k I mean is asking how many escenarios only men's get elected for office, tell me if I am wrong, please review my results in the thread to tell if I am right Whit my calculations of probabilities. Of a and c
    For B I get it culd be done Whit pairs but as you said there is no need but, should i permute 3 of 1 being the women's number? But if I permute then what how to calculate that probability?
     
  10. Jun 18, 2017 #9
    My answer to b is 3/7 ?
     
  11. Jun 18, 2017 #10
    Update answer I ended Whit 6/21 pairs of woman elected president, so this gives me the same probabilities of men elected to both offices, is this right?
     
  12. Jun 18, 2017 #11

    Ray Vickson

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    No. The probability that a woman is elected president is not equal to the probability that both electees are men. I am not allowed to tell you the answer, but I can tell you that 6/21 is wrong for part (b) (but is correct for part (a)). For part (a) I looked at permutations (not combinations) and got the same answer as you got. I will let you figure out why that is.
     
  13. Jun 18, 2017 #12
    I think I forgot to add the other 3 pairs of women elected for both offices, so it must be 9/21 this is suspicious because
    Of I do the 3/7 gets the same result of probability so it must be right, correct me if not please
     
  14. Jun 19, 2017 #13

    haruspex

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    That's right. Each route, when correct, gets to 3/7.
     
  15. Jun 20, 2017 #14
    then is solved many thanks
     
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