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# Find the probability that a 4 person committee chosen at ran

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Intern
Joined: 14 Dec 2009
Posts: 17
Find the probability that a 4 person committee chosen at ran  [#permalink]

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14 Dec 2009, 21:05
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0% (00:00) correct 100% (00:42) wrong based on 16 sessions

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Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains:

A) exactly 1 woman
B) at least 1 woman
C) at most 1 woman
Manager
Joined: 09 May 2009
Posts: 147

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14 Dec 2009, 23:11
5
A) 7C1* 11C3/ 18C4
B) 1 - (11C4/18C4)
C) (11C4/18C4) + (7C1*11C3/18C4)
##### General Discussion
SVP
Joined: 06 Sep 2013
Posts: 1522
Concentration: Finance

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27 Feb 2014, 09:46
Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains:

A) exactly 1 woman
B) at least 1 woman
C) at most 1 woman

Let's see here's what I got for this one:

A) W (NW) (NW)

3C2* (7/18)(11/17)(10/16)

B) 1 - P (NW)
1 - (11/18)(10/17)(9/6)

C) At most 1 woman is either none or 1 women hence

(11/18)(10/17)(9/16) + (7/18)(11/17)(10/16)

Bunuel does this look OK? I'm specially concerned about the first case. Do we need to 3C2 for order of them? Cause we could have W (M) (C) in which case we would have 3! instead of 3C2.

Many thanks!
Cheers
J

Bumpingggg
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Joined: 02 Aug 2009
Posts: 8335
Re: Find the probability that a 4 person committee chosen at ran  [#permalink]

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22 Jun 2016, 09:30
2
delta09 wrote:
Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains:

A) exactly 1 woman
B) at least 1 woman
C) at most 1 woman

Hi,

the way you should work on each is..
6M, 7W and 5C.... 4 person committe to be made..

A) exactly 1 woman
Choose 1 women out of 7 = 7C1..
choose rest 3 out of 6M and 5C = 11C3..
ans $$= 7C1*11C3..$$

B) at least 1 woman
Total ways = $$(6+7+5)C4 = 18C4$$...
ways NO women is choosen = $$(6+5)C4 = 11C4$$...
ways atleast ONE women is choosen = $$18C4 - 11C4$$...

C) at most 1 woman
So 1 women or no women..
No women = 11C4..
1 women (from A) = 7C1*11C3..
ans = $$11C4+7C1*11C3$$
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Re: Find the probability that a 4 person committee chosen at ran  [#permalink]

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07 Dec 2019, 23:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Find the probability that a 4 person committee chosen at ran   [#permalink] 07 Dec 2019, 23:58
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