Question

Suppose that 7 female and 5 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:

(a) 350

(b) 175

(c) 21

(d) 196

Step-by-step explanation:

Number of females = 7

Number of males = 5

Total ways of selecting r items from n items is

$^nC_r=\dfrac{n!}{r!(n-r)!}$

(a)

Total ways of selecting 3 females and 2 males.

$\text{Total ways}=^7C_3\times ^5C_2$

$\text{Total ways}=\dfrac{7!}{3!(7-3)!}\times \dfrac{5!}{2!(5-2)!}$

$\text{Total ways}=35\times 10$

$\text{Total ways}=350$

(b)

Total ways of selecting 4 females and 1 male.

$\text{Total ways}=^7C_4\times ^5C_1$

$\text{Total ways}=32\times 5$

$\text{Total ways}=175$

(c)

Total ways of selecting 5 females.

$\text{Total ways}=^7C_5\times ^5C_0$

$\text{Total ways}=21\times 1$

$\text{Total ways}=21$

(d)

Total ways of selecting at least 4 females.

Total ways = 4 females + 5 females

$\text{Total ways}=175+21$

$\text{Total ways}=196$