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PS-Probability

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Senior Manager
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PS-Probability [#permalink] New post 02 Dec 2008, 04:20
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There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

How many possible ways are there to solve this question ?
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Re: PS-Probability [#permalink] New post 02 Dec 2008, 08:51
1. 6C3

Ben is already chosen need 3 more committe members

Kelly not chosen . Assume she is not there. Hence total member left is 6
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Re: PS-Probability [#permalink] New post 02 Dec 2008, 11:45
walker wrote:
Why not (2C1 * 6C3) / (8C4) ?


walker, unless I am missing something, Ben is to be selected and Kelly is not to be selected. Hence, for Ben, 1C1.
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Re: PS-Probability [#permalink] New post 02 Dec 2008, 11:57
Expert's post
I think we should take into account two possible cases: (Ben,x,x,x)(Kelly,x,x,x) and (Kelly,x,x,x)(Ben,x,x,x)
Therefore, we need chose Ben or Kelly for first committee (2C1) and three others not Ben and not Kelly (3C6)
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Re: PS-Probability [#permalink] New post 02 Dec 2008, 12:18
Walker,

Great explanation. I also got the same result as scthakur got. But I see now why I should use 2c1 instead of 1c1...

Thanks.
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Re: PS-Probability [#permalink] New post 03 Dec 2008, 04:45
walker wrote:
Why not (2C1 * 6C3) / (8C4) ?


Because Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.
so the total is (1c1.6c3)/8c4 which is 2/7
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Re: PS-Probability [#permalink] New post 03 Dec 2008, 05:48
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Yeah, you are right. 2C1 would be correct if we had 2 committees but here we have only one committee.
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Re: PS-Probability [#permalink] New post 03 Dec 2008, 08:35
walker wrote:
I think we should take into account two possible cases: (Ben,x,x,x)(Kelly,x,x,x) and (Kelly,x,x,x)(Ben,x,x,x)
Therefore, we need chose Ben or Kelly for first committee (2C1) and three others not Ben and not Kelly (3C6)


Why are you considering (Kelly, x, x, x)? Kelley should be excluded.
(Ben,x,x,x) is fine.
so the prob = 6c3 / 8c4

Quote:
There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

How many possible ways are there to solve this question ?

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Re: PS-Probability   [#permalink] 03 Dec 2008, 08:35
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