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A committee of 3 has to be formed out of 6 people. John and

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BG
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A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 09:04
A committee of 3 has to be formed out of 6 people. John and Viki are among these. What is the prob that John will be selected but Viki not?



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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 11:31
Total ways = C(6,3) = 20.
Prob that John will be selected but Viki would not is = 6
[John is selected and Viki not selected = Select 2 people from 4 = 6]

Thus probability = 6/20 = 3/10.

If the statement means that "Both John and Viki are not selected" should be included in the scenario, then to the favorable scenario we can add C(4, 3) = 4, giving a probability of 1/2. However, I don't think that scenario can be included
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 11:45
i think you should multiply 6 arrangements of John and not Viki by 3 since John is there and he can sit in each of the 3 seats.

6*3/20=9/20
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 12:38
4c2/6c3 = 6/20 = 3/10
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 13:12
july05 wrote:
why is it just 4c2?


choose 2 out of 4 (or 4 choose 2)

4!/(4-2)!2!
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 13:24
but what about the 3rd person who can also switch seats?
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 13:30
july05 wrote:
but what about the 3rd person who can also switch seats?


in this problem order doesn't matter, and combinations or 'choosing' ignore the order. I am not sure what do you mean by switching seats,

but the logic behid solving the problem is the following: 3 people can be chosen out of 6
in 6c3 ways

now assume Viki is already not selected and Jon is already selected, so you are left with 4 people to fill two extra spots on the committee, so you have 4c2
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   02 Jun 2005, 13:38
i got it. i guess i read so many problems in the last hour that i thought that they were supposed to be seated. and this problem is about selecting a group.... :roll:
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Re: A committee of 3 has to be formed out of 6 people. John and [#permalink] New post   03 Jun 2005, 11:11
sparky wrote:
4c2/6c3 = 6/20 = 3/10


that one has my vote
0.3



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