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

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Senior Manager
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Six Married couples are standing in a room. If 4 people are [#permalink] New post 17 Dec 2007, 20:42
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.
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Re: Probability questions.... [#permalink] New post 17 Dec 2007, 20:59
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
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Re: Probability questions.... [#permalink] New post 23 Dec 2007, 03:26
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 .
Can anybody please explain ?
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 [#permalink] New post 23 Dec 2007, 12:24
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
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 [#permalink] New post 23 Dec 2007, 12:56
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
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 [#permalink] New post 23 Dec 2007, 14:59
You are right, but you have to distinguish among all possible player.
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Re: Probability questions.... [#permalink] New post 23 Dec 2007, 15:26
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 .
Can anybody please explain ?


Thats a good one TIGER.

dynamo:
Divide each of the numbers given by TIGER by 12C4
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Re: Probability questions.... [#permalink] New post 19 Jun 2008, 04:13
my answers are:

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
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Re: Probability questions.... [#permalink] New post 19 Jun 2008, 04:13
my answers are:

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
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Re: Probability questions.... [#permalink] New post 19 Jun 2008, 05:21
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, 05:21
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