Find the probability that a 4 person committee chosen at ran : Problem Solving (PS)

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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: 18

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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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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Manager

Joined: 09 May 2009

Posts: 160

Re: probabilty-3q [#permalink]

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14 Dec 2009, 23:11

A) 7C1* 11C3/ 18C4
B) 1 - (11C4/18C4)
C) (11C4/18C4) + (7C1*11C3/18C4)
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Joined: 06 Sep 2013

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Concentration: Finance

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Re: probabilty-3q [#permalink]

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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

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

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22 Jun 2016, 09:30

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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30 Jan 2018, 04:44

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