Question

Personnel selection. Suppose that 6 female and 6 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 12 ​finalists, what is the probability of selecting ​(A) 3 females and 2​ males? ​(B) 4 females and 1​ male? ​(C) 5​ females? ​(D) At least 4​ females?

Answer 1

Answer:

The answer is below

Step-by-step explanation:

The possible ways of choosing r items from a total of n item is given as C(n,r) = $\frac{n!}{(n-r)!r!}$

Suppose that 6 female and 6 male applicants have been successfully screened for 5 positions.

Since they are 12 finalists and only 5 positions, the total number of possible outcomes is:

$Total\ number\ of \ outcome (n(s))=C(12,5)=\frac{12!}{(12-5)!5!} =792$

A)  3 females and 2​ males

The possible ways of choosing 3 females and 2​ males is given as:

$n(e)=C(6,3)*C(6,2) = \frac{6!}{(6-3)!3!}* \frac{6!}{(6-2)!2!}=20*15=300$

probability of selecting 3 females and 2​ males P(E) = $\frac{n(e)}{n(s)}=\frac{300}{792}=0.379$

B)  4 females and 1​ males

The possible ways of choosing 4 females and 1 males is given as:

$n(e)=C(6,4)*C(6,1) = \frac{6!}{(6-4)!4!}* \frac{6!}{(6-1)!1!}=15*6=90$

probability of selecting 4 females and 1 males P(E) = $\frac{n(e)}{n(s)}=\frac{90}{792}=0.1136$

C)  5 females

The possible ways of choosing 5 females  is given as:

$n(f)=C(6,5) = \frac{6!}{(6-5)!5!}=6$

probability of selecting 5 females = $\frac{n(f)}{n(s)}=\frac{6}{792}$

​(D) At least 4​ females

This can be done by choosing 4 females out of 6 females i.e. C(6,4) , 5 females out of 6 females i.e C(6,5) and 1 male out of six male i.e (6,1)

The possible ways of choosing atleast 4 females  is given as:

$n(h)=C(6,5)*C(6,4)*C(6,1)= \frac{6!}{(6-5)!5!}* \frac{6!}{(6-4)!4!}* \frac{6!}{(6-1)!1!}=6*15*6=540$

probability of selecting atleast 4 females = $\frac{n(h)}{n(s)}=\frac{540}{792}$