Q:

# 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 selectinga. 3 females and 2 males?b. 4 females and 1 male?c. 5 females?d. at least 4 females?

Accepted Solution

A:
Answer:(a) 350(b) 175(c) 21(d) 196Step-by-step explanation:Number of females = 7Number of males = 5Total 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$$\text{Total ways}=175+21$$$$\text{Total ways}=196$$