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Probability, There are 20 laborers to four different construction jobs

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Please help me in my attempt:

    There are 20 laborers to four different construction jobs including 4 workers of ethnic minority. The first job required 6 laborers;
    the second, third, and fourth utilized 4, 5, and 5 laborers, respectively. What is the probability that:

    a) an ethnic group member is assigned to each type of job?
    b) no ethnic group member is assigned to a type 4 job?


    3. The attempt at a solution
    a) [16!/(5!3!4!4!)]/[20!/(6!4!5!5!)]

    equals to 0.0051. Thus, there is 0.5% chance that each minority would be assigned to each job


    b) I offer one way:

    1- [16!/(6!4!5!1!)] / [20!/(6!4!5!5!)]



    Please offer another way.
     
  2. jcsd
  3. Aug 28, 2012 #2

    vela

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    Re: Probability

    Can you explain your logic behind your answer to part (a)? It may be obvious to you since you're in the class, but I don't see where your expression came from off the top of my head.
     
  4. Aug 28, 2012 #3
    Re: Probability

    OK - Now I am correcting myself a bit. The logic behind part a answer:

    The total number of all the possible options is [20!/(6!4!5!5!)].
    Now I know that I have to put 4 minority workers into each job. So I have 4! options of distribution of those minorities. I also have 16! workers that are left to work with the rest of the jobs. So it goes:

    (4!16!)/(5!3!4!4!)
     
  5. Aug 28, 2012 #4

    vela

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    Re: Probability

    That comes out to be greater than 1.
     
  6. Aug 28, 2012 #5
    Re: Probability

    Then, the final answer for part a is:
    (4!16!)/(5!3!4!4!) divided by [20!/(6!4!5!5!)]
     
  7. Aug 28, 2012 #6

    vela

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    Re: Probability

    Oh, OK. That looks right to me.
     
  8. Aug 28, 2012 #7

    vela

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    Re: Probability

    For part (b), what do you mean when you say a type 4 job?
     
  9. Aug 28, 2012 #8
    Re: Probability

    It means the very last job, the fourth one.
     
  10. Aug 28, 2012 #9

    vela

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    So (b) isn't a complementary event to that of (a), so the method you tried isn't correct.
     
  11. Aug 28, 2012 #10

    Ray Vickson

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    I get a totally different answer for (a). We can (randomly) assign workers to jobs sequentially; that is, first assign job 1, then job 2, etc. For job 1 there are N1 = 4 minority people and N2 = 20-4 = 16 non-minorities. The probability of assigning k1 minorities to job 1 is the hypergeometric distribution with parameters (N1,N2,n) = (4,16,6):
    [tex] p_1(k_1) = \frac{ {N_1 \choose k_1} {N_2 \choose n-k_1}}{{N_1 + N_2 \choose n}},[/tex] where we use [itex] N_1 = 4,\; N_2 = 16,\; n = 6, \; k_1 = 1.[/itex] Now, given that job 1 has one minority member, we job 2 sees the situation with [itex]N_1 = 3, N_2 = 11, n = 4, [/itex] so [itex]p_2(k_2) \equiv P\{ k_2 | k_1 \}[/itex] is given by the hypergeometric distribution with the new parameters. Update again to get the probability that job 3 gets one minority, given that jobs 1 and 2 each have one, etc. Now, of course, the overall probability is the product of the individual probabilities, since [tex] P\{ k_1=1, k_2=1, k_3=1, k_4=1 \} = P\{k_1=1\} \cdot P\{ k_2=1 | k_1 = 1 \} \cdot P\{k_3 = 1 | k_1=1, k_2=1 \}
    \cdot P\{ k_4 = 1 | k_1=1, k_2=1, k_3=1\}.[/tex]

    RGV
     
    Last edited: Aug 28, 2012
  12. Aug 28, 2012 #11
    [QUOTE=Ray Vickson}


    I later corrected myself. Please see if my solution that follows the original posting is correct?

    Thank you!
     
  13. Aug 29, 2012 #12

    Ray Vickson

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    Last edited: Aug 29, 2012
  14. Aug 29, 2012 #13
    Thanks, this is correct!
     
  15. Aug 29, 2012 #14

    Ray Vickson

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    So (he repeats): what did you get for part (b)?

    RGV
     
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