Combination problem

[adamz]

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

[GmatMathPro]

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

[VivianKerr]

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.

[user123321]

total no of committees possible = 6c3/2! = 20/2 = 10
committes formed with concerned two members = 4c1

=4/10*100 = 40%

user123321

[Cheese12]

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 %

[GMATGuruNY]

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.