Question 1118015: What is the probability if you randomly select 6 people from 7 men and 7 women that will have at least 3 women in the group?
The answer my teach gave us is 302/429 or 0.704
I've tried many ways but I can't figure out how to get this answer. Thank you!
Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! You should post what you've tried, so a tutor can see where you are getting stuck.
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)
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!)
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.
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)
Answer by ikleyn(52813)  (Show Source):
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