Question 1118015
You should post what you've tried, so a tutor can see where you are getting stuck.<br>

By considering the complementary problem (2 or fewer women), the problem is computationally easier:
P(3 or more women are chosen) = 1 - P(2 or less women are chosen) <br>

P(exactly 0 women are chosen) = (C(7,0)*C(7,6)) / C(14,6)
P(exactly 1 woman is chosen) =  (C(7,1)*C(7,5)) / C(14,6)
P(exactly 2 women are chosen) = (C(7,2)*C(7,4)) / C(14,6)


Where C(n,r) = n! / ((n-r)!*r!) <br>

Compute the above three values, add them all together, then subtract the result from 1 to arrive at 0.704.  
[ If you work with the fractions, note that the denominator will be 3003 (= 7*429) ] 

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Tutor ikleyn has missed this: for each selection of three or more women, there are N>0 selections of men that increase the numerator in a multiplicative fashion.  For example, W1,W2,W3,M1,M2,M3  is different than W1,W2,W3,M1,M2,M4  even though the same three women are selected in both cases.  <br>

The numerator in fact should be  C(7,3)*C(7,3) + C(7,4)*C(7,2) + C(7,5)*C(7,1) + C(7,6)*C(7,0)  = 2114.    These represent the number of ways of selecting 3 women and 3 men, plus 4 women and 2 men,  etc.
 
There you have your teacher's fraction, as  2114/3003 = (7*302) / (7*429)