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A certain consulting firm employs 8 men and 4 women. In

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SVP
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A certain consulting firm employs 8 men and 4 women. In [#permalink]

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New post 11 Dec 2005, 20:08
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18. A certain consulting firm employs 8 men and 4 women. In March, 3 employees are selected at random to represent the company at a convention, what is the probability that the representatives will NOT all be men?
(A)14/55
(B)3/8
(C)41/55
(D)2/3
(E)54/55

Last edited by Bhai on 12 Dec 2005, 08:01, edited 1 time in total.

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New post 11 Dec 2005, 20:35
Bhai wrote:
18. A certain consulting firm employs 8 men and 4 women. In March, 3 employees are selected at random to represent the company at a convention, what is the probability that the representatives will NOT all be men?
(A)55/14
(B)8/3
(C)55/41
(D)3/2
(E)55/54
May be I am getting something wrong but how come the prob can be >1 ?

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New post 11 Dec 2005, 20:37
12C3 = 220
8C3 = 56

P(all men) = 56/220=14/55
P(! all men) = 1-14/55 = 41/55

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New post 11 Dec 2005, 20:40
ywilfred wrote:
12C3 = 220
8C3 = 56

P(all men) = 56/220=14/55
P(! all men) = 1-14/55 = 41/55

Agree same reasoning But the answer choices?May be a typo?

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New post 12 Dec 2005, 08:02
This one is also edited.

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New post 12 Dec 2005, 08:20
(C) looks like the more attractive Lucky Twin.

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New post 12 Dec 2005, 15:18
I wasn't sure how to do this, so i tried this little trick:

Look for the two answer choices that add up to and pick the one that the qn refers to. If all were men, it is the lesser probability, if all weren't men, pick the larger.

Chose 41/55 with this strategy.

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New post 12 Dec 2005, 15:37
Hi Folks ,

I got C by

1- (8/12*7/11*6/10) = (1-14/55) = 41/55

Can someone please clear one doubt I have on this.

Why isnt Binomial Distrubution Applicable here. I tried to get the same result using p=2/3 and applying Binomial Distribution but did not get the ans.

Whats wrong with that approach ?

thanks

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  [#permalink] 12 Dec 2005, 15:37
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A certain consulting firm employs 8 men and 4 women. In

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