Randomly choosing job applicants

[nickar1172]

Among 15 male applicants for a job, 9 have wives. If 3 of the males are randomly chosen for consideration, find the probability that:

a) none have wives
b) only 1 has a wife
c) 2 have wives
d) all 3 have wives

is this hypergeometric distribution and if so do are each of the choices (a,b,c,d) represent the value for n (not N which is the population size)?

 

[zzephod]

Among 15 male applicants for a job, 9 have wives. If 3 of the males are randomly chosen for consideration, find the probability that:

a) none have wives
b) only 1 has a wife
c) 2 have wives
d) all 3 have wives

is this hypergeometric distribution and if so do are each of the choices (a,b,c,d) represent the value for n (not N which is the population size)?

Yes this is a hypergeometric distribution question with population size $$N$$ being $$15$$, total successes in the population $$K$$ being $$9$$ , sample size $$n$$ being $$3$$ and the number of successes $$k$$ being $$0,\ 1,\ 2,\ 3$$ for parts a, b, c and d respectively.

See the Wikepedia article

However having made the observation that this question involves the hypergeometric distribution, with these numbers I would do these calculations by hand:

a) First choice has no wife with prob $$6/15$$, second $$5/14$$, third $$4/13$$ so required prob $$\frac{6\times 5 \times 4}{15 \times 14 \times 13}$$

b) First choice has a wife with prob $$9/15$$, second has no wife $$6/14$$, third has no wife $$5/13$$ so the prob $$\frac{9\times 6 \times 5}{15 \times 14 \times 13}$$ but any of the choices may have the wife so the final answer is \(3\) times this.

c) ....

 

[HallsofIvy]

Among 15 male applicants for a job, 9 have wives. If 3 of the males are randomly chosen for consideration, find the probability that:

a) none have wives
b) only 1 has a wife
c) 2 have wives
d) all 3 have wives

is this hypergeometric distribution and if so do are each of the choices (a,b,c,d) represent the value for n (not N which is the population size)?

Yes, it is a hypergeometric distribution but I would not consider that particularly relevant to solving this problem. That it is "selection without replacement" is much more relevant.

I don't know what you mean by "represent the value for n" since you haven't said what "n" is or what it represents.

Here is how I would do this problem:
a) Initially there are 15 applicants, 9 of whom have wives, 6 do not. The probability the first chosen does not have a wife is 6/15= 2/5. There are now 14 applicants, 5 of whom do not have wives. The probability the second chosen does not have a wife is 5/14. There are now 13 applicants, 4 of whom do not have wives. The probability the third chosen does not have a wife is 4/13.

The probability none of the three have wives is (2/5)(5/14)(4/13)= 8/((13)(14))

b) Start similarly. Initially, there are 15 applicants, 9 of whom have wives. The probability the first applicant chosen has a wife is 9/15= 3/5. There are now 14 applicants, 6 of whom do NOT have a wife. The probability the next chosen does not have a wife is 6/14= 3/7. Finally, there are 13 applicants, 5 of whom do not have wives. The probability the last chosen does not have a wife is 5/13.

The probability the first chosen has a wife and the other two do not is (3/5)(3/7)(5/13)= 9/((7)(13)).

Of course, you also have to consider the cases in which the second chosen or the third chosen is the one with the wife but it is easy to see that, while the fractions are different, the numerators and denominators are the same, just in different order, so the product is the same. To get the total probability, multiply that fraction by 3: 27/((7)(13)).

(c) and (d) are the same, just with "has a wife" and "does not have a wife" reversed.