Question

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?

Answer 1

0.4196

0.2098

0.0280

0.2378

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