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probability

Discussion in 'DAT Discussions' started by BodybldgDoc, Apr 25, 2007.

  1. BodybldgDoc

    BodybldgDoc Guest

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    Jan 13, 2007
    In your closet
    How do we figure this one out?

    What will be the likelihood of having exactly 1 boy in a family planning for three children?

     
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  3. DailyDrivenTJ

    DailyDrivenTJ 2+ Year Member

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    Mar 20, 2007
    Permutation of having 3 children: 2x2x2=8

    There are 3 ways to have exactly one boy in 3 children.

    BGG
    GBG
    GGB

    So I am seeing 3/8.

    What is the answer?
     
  4. Lonely Sol

    Lonely Sol cowgoesmoo fan! 10+ Year Member

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    Dec 13, 2005
    Atlanta, GA
    3!/2!1! = 3 (since order doesnt matter)

    1/2^3 = 1/8

    So, 3/8 is the probablity of having 1 boy out of 3 tries.
     
  5. BodybldgDoc

    BodybldgDoc Guest

    712
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    Jan 13, 2007
    In your closet
    the answer is 3/8. thanks guys
     
  6. Greensalad

    Greensalad 10+ Year Member

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    Apr 16, 2007
    Can you please explain why you did 3!/2!1!?

    I can't find a good permutations website.
     
  7. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

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    Oct 25, 2006
    NJ
    (3 choose 1) = # of ways to select 1 object out of 3 (without regard to order).

    (n choose k) = n! / [(n-k)!*k!]

    So n! would represent the # of ways to select all 3 objects in any order. When you divide by (n-k)! you are cutting it off after k objects. So instead of considering n objects to select, you consider the first k objects.

    Ex: 6 choose 2. The n! = 6! means you want to select all 6 objects. But the (n-k)! = (6-2)! = 4! means you only consider the first k = 2 objects: 6!/4! = 6*5. There are 6 choices for the first selection and 5 for the second selection. Then you stop.

    The 1/k! accounts for the # of ways to order the objects. With k objects you can order them k! ways (there are k ways to choose which comes first, k-1 ways to choose which comes second, etc). So if you don't care about order, you divide by k! because that's how many times each arrangement is 'duplicated' in your selection process, so to speak.

    Ex: If you select 3 objects from 10, you could pick A B C or B C A or B A C to name a few. But those are the same 3 objects. Since there are 6 ways to arrange the letters A, B, and C, you divide by 6 (3!) to account for this. Only 1 in every 6 selections will be a different group of 3.
     
  8. DASicari

    DASicari 10+ Year Member

    180
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    Mar 25, 2007
    Orange County, CA
    :scared:

    I hope I learn this in statistics in the fall.
     
  9. Greensalad

    Greensalad 10+ Year Member

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    Apr 16, 2007
    So what if the OP cared about order? Could you please show me by incorporating it into his question? Thanks. Mr. 28 QR.
     
  10. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
    5
    Oct 25, 2006
    NJ
    You just don't divide by k!. Instead of n!/[(n-k)!*k!] you have n!/(n-k)!

    So it would be 3!/2! = 3 still... and it's the same because you are only looking for ways to select one kid to be a boy out of the three... and there is only one way to order one object.

    If you needed something like 5 items out of 12 with and without order mattering, then you'd have:

    (12 choose 5) = 12!/7!5! = 792 ways to choose without regard to order.
    And then 12!/7! = 95,040 ways to choose if order matters.

    In this case you are selecting 5 objects. If order matters then you could choose A, B, C, D, E or A, C, D, E, B or D, B, A, E, C etc. How many ways can you arrange 5 objects? You can pick any of the 5 to go first, then 4 to go second, 3 to go third, 2 to go fourth, and 1 to go last. That's 5! = 120 ways. So I can select objects A B C D and E in 120 different ways. If order matters then that's fine - we want to consider those separate ways of choosing them. But if order doesn't matter, we only need to consider 1 out of those 120 ways. That's why you divide 95,040 by 120 (or 5!) and get 792.
     
  11. Greensalad

    Greensalad 10+ Year Member

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    Apr 16, 2007
    Cool, thanks!

    Ok. What if, instead of having exactly 1 boy, the family has at least 1 boy?
     
  12. DailyDrivenTJ

    DailyDrivenTJ 2+ Year Member

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    Mar 20, 2007
    7/8.

    At least 1 boy means, not getting all three girls.

    There are 8 possibilities of "making" children but only 1 way to make all 3 girls.

    8-1 gives you 7, so it is 7 out of 8.
     
  13. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
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    Oct 25, 2006
    NJ
    Then that's the probability of not having 0 boys. The probability of having 0 boys is (3 choose 0) * (1/2)^3 or 1/8. Then since you want the probability of this not happening, it's 1 - (1/8) = (7/8).
     
  14. Greensalad

    Greensalad 10+ Year Member

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    Apr 16, 2007
    So then what would it be to have at least 2 boys out of 3 tries?

    Thanks :)
     
  15. jeezy

    jeezy 2+ Year Member

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    Mar 23, 2007
    P(2 boys) + P(3 boys)

    P(2) = 3!/2!1! * (1/2)^3 = 3/8
    P(3) = 1/8

    so 1/2
     
  16. 113zami

    113zami 2+ Year Member

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    Oct 4, 2007
    I don't get the explanation to that last question, why are you adding? does adding account for the "at least" part of the question

    thanks
     
  17. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
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    Oct 25, 2006
    NJ
    Yeah since P(2 or 3) = P(2) + P(3) because they are independent events.
     
  18. 151AND8TH

    151AND8TH 2+ Year Member

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    Dec 7, 2006
    NEW YORK, NY
    the binomial expension I use for most of these "kids" problem
    since you can only have a boy or a girl.. the probability of having a girl is always 1/2. the probability of having a boy is also always 1/2
    a= girl; b=boy

    example
    2 girl 3 boy total 5 kids
    (a+b)^5= (a^5)+5(a^4b)+10(a^3b^2)+10(a^2b^3)+5(ab^4)+b^5


    so you would select 10(a^2b^3) and replace both (a) and (b) with 1/2


    ..... Hence for a set of kids with 1 being a boy then 2 must be girls you would expend
    (a+b)^3= (a^3)+3(a^2b)+3(ab^2)+(b^3)

    you would use this part of the expension: 3(a^2b)
    and replacing (a) and (b) with 1/2 you would end up with : 3/8 as athe probablity of having 1 boy and 2 girls

    or you could use P=(n!/s!t!)a^s*b^t
    where (s) is the number of girl and (t) the number of girl and :thumbdown: the total number of kids
    which would also give you 3/8

    BUT

    when it comes to at "least x kid" problems these equations seem to be useless.. so what are your take on this matter Streetwolf? are these genetics equations really useless?
     

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