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) = [tex]\frac{n!}{(n-r)!r!}[/tex]
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:
[tex]Total\ number\ of \ outcome (n(s))=C(12,5)=\frac{12!}{(12-5)!5!} =792[/tex]
A) 3 females and 2 males
The possible ways of choosing 3 females and 2 males is given as:
[tex]n(e)=C(6,3)*C(6,2) = \frac{6!}{(6-3)!3!}* \frac{6!}{(6-2)!2!}=20*15=300[/tex]
probability of selecting 3 females and 2 males P(E) = [tex]\frac{n(e)}{n(s)}=\frac{300}{792}=0.379[/tex]
B) 4 females and 1 males
The possible ways of choosing 4 females and 1 males is given as:
[tex]n(e)=C(6,4)*C(6,1) = \frac{6!}{(6-4)!4!}* \frac{6!}{(6-1)!1!}=15*6=90[/tex]
probability of selecting 4 females and 1 males P(E) = [tex]\frac{n(e)}{n(s)}=\frac{90}{792}=0.1136[/tex]
C) 5 females
The possible ways of choosing 5 females is given as:
[tex]n(f)=C(6,5) = \frac{6!}{(6-5)!5!}=6[/tex]
probability of selecting 5 females = [tex]\frac{n(f)}{n(s)}=\frac{6}{792}[/tex]
(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:
[tex]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[/tex]
probability of selecting atleast 4 females = [tex]\frac{n(h)}{n(s)}=\frac{540}{792}[/tex]
