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Probability, There are 20 laborers to four different construction jobs

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Homework Statement


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?


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.
 

Answers and Replies

  • #2
vela
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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.
 
  • #3
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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!)
 
  • #4
vela
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That comes out to be greater than 1.
 
  • #5
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Then, the final answer for part a is:
(4!16!)/(5!3!4!4!) divided by [20!/(6!4!5!5!)]
 
  • #6
vela
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Oh, OK. That looks right to me.
 
  • #7
vela
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For part (b), what do you mean when you say a type 4 job?
 
  • #8
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It means the very last job, the fourth one.
 
  • #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.
 
  • #10
Ray Vickson
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Homework Statement


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?


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.
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
 
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  • #11
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[QUOTE=Ray Vickson}


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

Thank you!
 
  • #12
Ray Vickson
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Ray Vickson} I later corrected myself. Please see if my solution that follows the original posting is correct? Thank you![/QUOTE said:
For part (a) I get answer = 40/323 [itex]\doteq[/itex] 0.123839.

What do you get for part (b)? I know you wrote something before, but are you sure that was the correct answer? Give reasons, don't just write things down.

RGV
 
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  • #13
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Thanks, this is correct!
 
  • #14
Ray Vickson
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Thanks, this is correct!
So (he repeats): what did you get for part (b)?

RGV
 

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