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Find the probability that a 4 person committee chosen at ran

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

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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
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Re: probabilty-3q [#permalink]

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New post 14 Dec 2009, 22:11
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A) 7C1* 11C3/ 18C4
B) 1 - (11C4/18C4)
C) (11C4/18C4) + (7C1*11C3/18C4)
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Re: probabilty-3q [#permalink]

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New post 27 Feb 2014, 08:46
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

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Re: Find the probability that a 4 person committee chosen at ran [#permalink]

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New post 22 Jun 2016, 08:30
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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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New post 30 Jan 2018, 03:44
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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] 30 Jan 2018, 03:44
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