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none of them are married

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by sanju09 » Sat Feb 21, 2009 3:32 am
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

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by DanaJ » Sat Feb 21, 2009 3:47 am
I'd say 1*10/11*8/10*6/9 = 16/33, but probabilities is not really my strong point.....
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by sureshbala » Sat Feb 21, 2009 3:49 am
Yes, my answer is 16/33.
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by DeepakR » Sun Feb 22, 2009 7:35 pm
Can someone please explain this ?

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Re: none of them are married

by Brent@GMATPrepNow » Sun Feb 22, 2009 9:23 pm
sanju09 wrote:Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

Produce your own response, links :)
One way is to treat this as a counting problem.
P(no couples) = (# ways to select 4 people with no couples)/(total number of ways to select 4 people)

Denominator: 12 people, choose 4. Can be accomplished in 12C4 ways

Numerator. First, select 4 couples from the 6 couples. This can be accomplished 6C4 ways. From each of the 4 selected couples, choose 1 member. There are
- 2 ways to select a member from the first couple
- 2 ways to select a member from the second couple
- 2 ways to select a member from the third couple
- 2 ways to select a member from the fourth couple

Total number of ways to select 4 people with no couples is 6C4x2x2x2x2

Probability = 6C4x2x2x2x2/12C4 = 16/33
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Re: none of them are married

by x2suresh » Mon Feb 23, 2009 10:15 am
sanju09 wrote:Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

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(12C1*10C1*8C1*6C1/4!)/12C4 = 12*10*8*6/12*11*10*9 = 48/99 = 16/33
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by welcome » Mon Feb 23, 2009 12:03 pm
12/12*10/11*8/10*6/9 = 16/33.
Shubham.
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