math
posted by Jon .
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 selected

Are they selected randomly?
If randomly, which would be an unusual way of selecting folks for training (think about that)..
a) 3/7 * 2/6 * 1/5
c) you can select those several ways..
m, m, w
m,w,m
w,m,m
so, probab=3/7*2/6*4/5 + 3/7*4/6*2/5 + 4/7*3/6*2/5 = amazing..3ways*4*3*2/7*6*5
You probably have some combination formulas for this in your text. 
I doubt order matters. And there is no formula given in the text. So I was thinking for A it would be 3/ 7 since there are 3 men out of the possible 7 chosen executives.

I do these by using combinations (since order does not matter)
e.g. #1
prob = C(3,3)/C(7,3) = 1/35
b) C(4,3)/C(7,3) = 4/35
c) C(3,2)*C(4,1)/C(7,3) = 3*4/35 = 12/35
d) C(3,1)*C(4,2)/C(7,3) = 3*6/35 = 18/35