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Combination problem

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by adamz » Thu Oct 27, 2011 7:59 pm
A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that John is a member of is Roger also a member?

10%
20%
30%
40%
50%

OA:D
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by GmatMathPro » Thu Oct 27, 2011 8:16 pm
For any subcommittee that John is in, we can choose the other two members in 5C2=10 ways. For any subcommittee that John AND Roger are on, we can choose the third member in 4 ways.

4/10=40%

D
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by VivianKerr » Thu Oct 27, 2011 8:17 pm
We need to find out how many of those committees will David + Michael be on together.

Committee #1: There is a 50% chance that David will be a member, leaving 2 options for the other 5 people. Michael's odds are 2/5 or 40%.

Chances of David and Michael being on the first committee toegther will be 50% x 40% = 20%.

Committee #2: Same logic here. Since we don't know which committee they'll be on, we add 20 + 20 = 40.

The answer is D.
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by user123321 » Thu Oct 27, 2011 8:31 pm
total no of committees possible = 6c3/2! = 20/2 = 10
committes formed with concerned two members = 4c1

=4/10*100 = 40%

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by Cheese12 » Fri Oct 28, 2011 3:08 am
total number of possibilities of subcommunities = 6!/3!3! = 20

since Roger and John are fixed...
we see the possiblities fr the rest (2 subcommunites) = 4!/3!1! = 4 x 2 = 8

8/20 * 100 = 40 %
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by GMATGuruNY » Fri Oct 28, 2011 3:40 am
adamz wrote:A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that John is a member of is Roger also a member?

10%
20%
30%
40%
50%

OA:D
Question rephrased: What is the probability that Roger is one of the other 2 people selected to be on the committee with John?

P(Roger is NOT selected) = 4/5 * 3/4 = 3/5.
Thus, P(Roger IS selected) = 1 - 3/5 = 2/5 = 40%.

The correct answer is D.
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