Posted by **Jon** on Wednesday, April 1, 2009 at 10:29am.

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

- math -
**bobpursley**, Wednesday, April 1, 2009 at 10:41am
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.

- math -
**Jon**, Wednesday, April 1, 2009 at 11:00am
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.

- math -
**Reiny**, Wednesday, April 1, 2009 at 1:39pm
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

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