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?

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) =

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:

A)  3 females and 2 males

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

probability of selecting 3 females and 2 males P(E) =

B)  4 females and 1 males

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

probability of selecting 4 females and 1 males P(E) =

C)  5 females

The possible ways of choosing 5 females  is given as:

probability of selecting 5 females =

(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:

probability of selecting atleast 4 females =

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step-by-step explanation:

step-by-step explanation:

yes it represnts a functio, their are not 2 of the same numbers on the both sides