4 women, 3 men, select 3
prob(all men) = C(3,3)/C(7,3) = 1/35
prob(all women) = C(4,3)/C(73,) = 4/35
prob(2men, 1 woman) = C(3,2)*C(4,1)/C(7,3) = 12/35
prob(2women, 1man) = C(4,2)*C(3,1)/C(7,3) = 18/35
Did you notice that all cases are covered and
1/35+4/35+12/35+18/35 = 1 ?
In a company there are 7 executives: 4 women and 3 men. 3 are selected to attend a management seminar. Find these probabilities. A) All 3 selected are men B) all 3 selected are women C) 2 men and 1 woman will be selected. D) 1 man and 2 woman will be
2 answers
I don't know it