Question

In a company, there are 7 executives: 4 women and 3 men. There are selected to attend a management seminar. Find the probabilities of
a) All men to be selected
b) 2 men and 1 woman will be selected
c) At least a man will be selected
d) At most 2 women will be selected

Answer 1

How many are you selecting?
From the choices I would conclude you are choosing 3 to form the group

a) prob(3 men) = C(3,3)*C(4,0)/C(7,3)
= 1/35

or Prob(3 m3n) = 3/7*2/6*1/5 = 6/210 = 1/35

b) Prob(2 men, 1 woman) = C(3,2)*C(4,1)/C(7,3)
= 3*4/35 = 12/35

or, could be MMW, MWM, or WMM
= 3/7 * 2/6 * 4/5 + 3/7 * 4/6 * 3/5 + 4/7 * 3/6 * 2/5)
= 4/35 + 4/35 + 4/35
= 12/35

c) exclude the case of all women
= 1 - C(4,3)/C(7,3)
=

d) let me know how you reasoned this one, and what you got