Suppose that 8 female and 6 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 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?

0.4196

0.2098

0.0280

0.2378

Step-by-step explanation:

Given that:

Number of females = 8

Number of males = 6

Total = 8 + 6 = 14

Number of positions = 5

Probability = required outcome / Total possible outcomes

Total possible outcomes = 14C5

nCr = n! / (n-r)! r!

14C5 = 2002 (from calculator)

(A) 3 females and 2 males?

Required outcome : 8C3 * 6C2 = 56 * 15 = 840

840 / 2002 = 0.4196

(B) 4 females and 1 male?

(8C4 * 6C1) / 14C5 = 420 / 2002 = 0.2098

(C) 5 females?

8C5 / 14C5 = 56 / 2002 = 0.0280

(D) At least 4 females?

4 female 1 male + 5 females

((8C4 * 6C1) + 8C5) / 14C5

(420 + 56) / 2002 = 0.2378

thats not a system of equations, that is a solution