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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
A) exactly 1 woman
B) at least 1 woman
C) at most 1 woman
14 Dec 2009, 23:11
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A) 7C1* 11C3/ 18C4
B) 1 - (11C4/18C4)
C) (11C4/18C4) + (7C1*11C3/18C4)
B) 1 - (11C4/18C4)
C) (11C4/18C4) + (7C1*11C3/18C4)
General Discussion
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,316
Own Kudos:
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
27 Feb 2014, 09:46
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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
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
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
