Algebra
posted by Jon .
From a group of 6 men and 4 women, a committee of 3 is to be selected at random. Find P(at least 2 women).
It might be a little out of order but this is what I got.
C(10,2)
(10*9*8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1*2*1)
=45
C(6,2)
(6*5*4*3*2*1)/(2*1)
=360
C(4,2)
(4*3*2*1)/(2*1
=12
360*12 = 4320
s/s+f
4320/4320+45
4320/4365
99% answer

no how about this
P(at least 2 women)=
P(exactly 2 women)+P(exactly 3 women)=
{(6!/(1!*5!))*(4!/(2!*2!))+
(4!/(3!*1!)}/{10!/(3!*7!)}=
40/120=1/3
answer = 1/3
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