A local union has ten members, four of whom are men. Three of the members are chosen by a lottery to represent the union on a bargaining council. Find the probability of each event.

Question

a) Two women and one man are chosen

b) At least one man is chosen

Answer 1

As there are 10 members, if 4 are men, 6 are women.

a) P(2 women and 1 man) = C(6,2)C(4,1)/C(10,3) = 15*4/120 = 1/2

Alternatively, you can input into the hypergeometric distribution on a computer 2 women from a sample of size 3 when 6 women in a population of 10 and that you are looking for the pmf (not cumulative).

=HYPGEOM.DIST(2,3,6,10,0)

b) probability at least one man is 1 - P(3 women) = 1 - C(6,3)/C(10,3) = 1 - 20/120 = 5/6

Alternatively, we could use =1 - HYPGEOM.DIST(3,3,6,10,0)