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# Six Married couples are standing in a room. If 4 people are

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Senior Manager
Joined: 20 Dec 2004
Posts: 251
Six Married couples are standing in a room. If 4 people are [#permalink]

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17 Dec 2007, 21:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.
d) at least one married couple is among the 4.
SVP
Joined: 29 Aug 2007
Posts: 2473

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17 Dec 2007, 21:59
neelesh wrote:
Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.
d) at least one married couple is among the 4.

a) P(2 married couples are chosen) = 6c2 = 15
b) p(no married couple is among the 4) = 12c4 - (6c1 x 10c2)= 495-270=225
c) p(exactly one married couple is among the 4) = (6c1 x 10c2) - 6c2 = 270-15=255
d) p(at least one married couple is among the 4) = (6c1 x 10c2) = 270
Manager
Joined: 16 Jan 2007
Posts: 73

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23 Dec 2007, 04:26
GMAT TIGER wrote:
neelesh wrote:
Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.
d) at least one married couple is among the 4.

a) P(2 married couples are chosen) = 6c2 = 15
b) p(no married couple is among the 4) = 12c4 - (6c1 x 10c2)= 495-270=225
c) p(exactly one married couple is among the 4) = (6c1 x 10c2) - 6c2 = 270-15=255
d) p(at least one married couple is among the 4) = (6c1 x 10c2) = 270

GMATTIGER.
The following question asks for a probability and probability can never be greater than 1 .
Director
Joined: 09 Jul 2005
Posts: 591

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23 Dec 2007, 13:24
p1=1/12*1/11*1/10*1/9
p2=1/12*10/11*8/10*6/9
p3=1/12*1/11
p4=1-p2
SVP
Joined: 28 Dec 2005
Posts: 1558

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23 Dec 2007, 13:56
automan wrote:
p1=1/12*1/11*1/10*1/9
p2=1/12*10/11*8/10*6/9
p3=1/12*1/11
p4=1-p2

automan, im with you on this, with one question ... why is your first probability 1/12 ? I had it just as 1, because for the first choice, anyone can be selected, as there are no restrictions
Director
Joined: 09 Jul 2005
Posts: 591

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23 Dec 2007, 15:59
You are right, but you have to distinguish among all possible player.
Manager
Joined: 21 May 2007
Posts: 120

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23 Dec 2007, 16:26
dynamo wrote:
GMAT TIGER wrote:
neelesh wrote:
Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.
d) at least one married couple is among the 4.

a) P(2 married couples are chosen) = 6c2 = 15
b) p(no married couple is among the 4) = 12c4 - (6c1 x 10c2)= 495-270=225
c) p(exactly one married couple is among the 4) = (6c1 x 10c2) - 6c2 = 270-15=255
d) p(at least one married couple is among the 4) = (6c1 x 10c2) = 270

GMATTIGER.
The following question asks for a probability and probability can never be greater than 1 .

Thats a good one TIGER.

dynamo:
Divide each of the numbers given by TIGER by 12C4
Manager
Joined: 28 Apr 2008
Posts: 110

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19 Jun 2008, 05:13

The main idea is that when we are dealing with probabilities of thstype of questions, the fact that order is relevant or not is irrelevant.

A) 6*5/(12*11*10*9)

B) 12*10*8*6/(12*11*10*9)=48/99

c) 6*10*8/(12*11*10*9)=4/99

D) 1-P(B)= 1-48/99=51/99
Manager
Joined: 28 Apr 2008
Posts: 110

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19 Jun 2008, 05:13

The main idea is that when we are dealing with probabilities of thstype of questions, the fact that order is relevant or not is irrelevant.

A) 6*5/(12*11*10*9)

B) 12*10*8*6/(12*11*10*9)=48/99

c) 6*10*8/(12*11*10*9)=4/99

D) 1-P(B)= 1-48/99=51/99
Manager
Joined: 10 Mar 2008
Posts: 67

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19 Jun 2008, 06:21
neelesh wrote:
Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.
d) at least one married couple is among the 4.

a)2 married couples are chosen
there are 6 married couples so 12 people
prob= 12*10/12c4=1/99

because if u select one person the prob to select its spouse is 1

b) no married couple is among the 4

12*10*8*6/12c4=16/33

c)exactly one married couple is among the 4.
1-((2 married couples are chosen)+(no married couple is among the 4))

1-(1/99+16/33)

d) at least one married couple is among the 4.

1-(no married couple is among the 4)
1-16/33
Re: Probability questions....   [#permalink] 19 Jun 2008, 06:21
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