There are 9 candidates total, 5 male and 4 female, who run for 4 committee positions. suppose that each person is equally likely to be elected. what is the probability that 2 males and 2 females are elected?

Use the hypergeometric distribution.
M=number of Men=5
F=number of women=4
m=number of men elected=2
f=number ow women elected=2.

Assuming equal chance to get elected, then
P(2M,2F)=C(M,m)*C(F,f)/C(M+F,m+f)
=C(5,2)*C(4,2)/C(9,4)
=10*6/126
=10/21

Reference: Hypergeometric distribution.

13/18

Step-by-step explanation:

6.5 · 1/9 = 6 5/10 · 1/96 5/10 · 1/9 = 6 1/2 · 1/96 1/2 · 1/9 = 13/2 · 1/913/2 · 1/9 = 13/18

Therefore, 6.5 · 1/9 = 13/18.