Probability, There are 20 laborers to four different construction jobs

[Kinetica]

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.

 

[vela]

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.

 

[Kinetica]

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!)

 

[vela]

That comes out to be greater than 1.

 

[Kinetica]

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

 

[vela]

Oh, OK. That looks right to me.

 

[vela]

For part (b), what do you mean when you say a type 4 job?

 

[Kinetica]

It means the very last job, the fourth one.

 

[vela]

So (b) isn't a complementary event to that of (a), so the method you tried isn't correct.

 

[Ray Vickson]

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):
$$p_1(k_1) = \frac{ {N_1 \choose k_1} {N_2 \choose n-k_1}}{{N_1 + N_2 \choose n}},$$ where we use $N_1 = 4,\; N_2 = 16,\; n = 6, \; k_1 = 1.$ Now, given that job 1 has one minority member, we job 2 sees the situation with $N_1 = 3, N_2 = 11, n = 4,$ so $p_2(k_2) \equiv P\{ k_2 | k_1 \}$ 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 $$P\{ k_1=1, k_2=1, k_3=1, k_4=1 \} = P\{k_1=1\} \cdot P\{ k_2=1 | k_1 = 1 \} \cdot P\{k_3 = 1 | k_1=1, k_2=1 \} \cdot P\{ k_4 = 1 | k_1=1, k_2=1, k_3=1\}.$$

RGV

 

[Kinetica]

[QUOTE=Ray Vickson}


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

Thank you!

 

[Ray Vickson]

For part (a) I get answer = 40/323 $\doteq$ 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

 

[Kinetica]

Thanks, this is correct!

 

[Ray Vickson]

Thanks, this is correct!

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

RGV