Posted by **ash** on Sunday, June 5, 2011 at 10:14pm.

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.

- math -
**MathMate**, Sunday, June 5, 2011 at 10:26pm
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)

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