Question

a class consists of 14 men and 16 women. A group of 5 is randomly chosen
a) the probability of this group containing at least 2 women and at least 2 men?
b) probability the group contains the same gender.

Answer 1

Total number of ways, N(all)
=30 choose 5
=C(30,5)
=30!/(5!(30-5)!)


A.
2 women+3 men
=16 choose 2 * 14 choose 3
=C(16,2)*C(14,3)
3 women+2 men
=16 choose 3 * 14 choose 2
=C(16,3)*C(14,2)
Number of ways to have at least two women and two men:
N1=C(16,2)*C(14,3)+C(16,3)*C(14,2)
Probability = N1/N(all)

B.
Number of ways for all men
=14 choose 5
=C(14,5)
Number of ways for all women
=16 choose 5
=C(16,5)
Number of ways for one gender
N2=C(14,5)+C(16,5)
Probability
=N2/N(all)