Hi,
6 members can be divided into two subcommittees of 3 members each in 6C3 = 20 ways
David and Michael can be in same group in 2 ways(either group1 or group2)
Now, the remaining member of the subcommittee can be selected from the remaining 4 in 4C1 = 4 ways
So, probability that David and Michael are in same subcommittee is 2*4C/20 = 2/5 = 40%

Hence, 4

No. of ways of forming a subcommittee = 6C3 = 20 (other committee will be formed automatically)

No. of committee's in which David and Michael are present = 1*1*4C1 = 4

Well i could not get why we need to multiply by 2. We found total no. of subcommittee and no. of subcommittee's in which David and Michael are present. How come this thing of gr1 / gr2 comes into picture. Our calculations should already include this because we did not take this into account while calculating.

To confirm this i wrote down all the cases. Lets say 6 the members are A,B,C,D,E,F then 20 possible subcommittees are
ABC
__D
__E
__F
ACD
__E
__F
ADE
__F
AEF
BCD
__E
__F
BDE
__F
BEF
CDE
__F
CEF
DEF

In any of these, any 2 members(say A & B) are together in only 4 cases.

saxenashobhit 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 Michael is a member of is David also a member?

(1)10%
(2)20%
(3)25%
(4)40%
(5)50%

OA 4

Please show steps[/spoiler]

Since the 6 people are being divided into groups, it is a given that Michael will be a member of one of the two groups.
Thus, we don't need to count the total number of possible groups; we need to determine only the probability that David is one of the 2 people chosen to be with Michael.
P(David is NOT chosen to be with Michael) = 4/5 * 3/4 = 3/5.
Thus, P(David IS chosen to be with Michael) = 1 - 3/5 = 2/5 = 40%.

The correct answer is D.

Hey Mitch How did you get 3/4?

winniethepooh wrote:Hey Mitch How did you get 3/4?

Two people must be assigned to Michael's committee.

P(first person chosen is not David):
Out of the 5 other people, 4 are not David.
P(first person chosen is not David) = 4/5.

P(second person chosen is not David):
Out of the 4 remaining people, 3 are not David.
P(second person chosen is not David): = 3/4.

Since for David not to be chosen both events must happen, we multiply:
P(David is not chosen to be with Michael) = 4/5 * 3/4 = 3/5.

Thus, P(David is chosen to be with Michael) = 1 - 3/5 = 2/5 = 40%.

GMATGuruNY wrote: Since the 6 people are being divided into groups, it is a given that Michael will be a member of one of the two groups.
Thus, we don't need to count the total number of possible groups; we need to determine only the probability that David is one of the 2 people chosen to be with Michael.
P(David is NOT chosen to be with Michael) = 4/5 * 3/4 = 3/5.
Thus, P(David IS chosen to be with Michael) = 1 - 3/5 = 2/5 = 40%.

The correct answer is D.

Hey Mitch,
If use a similar approach as that of yours, then i get the following.
Prob. that Michael is a member of the group that has David =1/5
Now the 3rd member can be any one of the left 4 members so their prob. = 4/4 =1

There group has prob. = 1/5*1 =1/5 .... Whats wrong with this method

vikram4689 wrote:No. of ways of forming a subcommittee = 6C3 = 20 (other committee will be formed automatically)

No. of committee's in which David and Michael are present = 1*1*4C1 = 4

Well i could not get why we need to multiply by 2. We found total no. of subcommittee and no. of subcommittee's in which David and Michael are present. How come this thing of gr1 / gr2 comes into picture. Our calculations should already include this because we did not take this into account while calculating.

To confirm this i wrote down all the cases. Lets say 6 the members are A,B,C,D,E,F then 20 possible subcommittees are
ABC
__D
__E
__F
ACD
__E
__F
ADE
__F
AEF
BCD
__E
__F
BDE
__F
BEF
CDE
__F
CEF
DEF

In any of these, any 2 members(say A & B) are together in only 4 cases.

Hi,
What about the bold cases:
If you select each of these as one subcommittee then the other subcommittee will be having AB together. So total 8.

Yeah got it, so you want to say that i consider AB when they were together in 1 group but similarly they can be in another group too. So multiply by 2.

vikram4689 wrote:Yeah got it, so you want to say that i consider AB when they were together in 1 group but similarly they can be in another group too. So multiply by 2.

Yup.. That should be the way because by saying all combinations in 6C3, we are counting in a similar way.

Thanks all. I understood Mitch approach. I have a question on going by combinations/counting method

On combination style - I have question on "all possible combination possible out of 6 people". I can select 3 people to make committee in 6C3 ways = 20 ways. Now these 3 people can join commitee 1 or committee 2. So total combinations should be - 2* 20 = 40 <this is what my question is - is this correct?>

Total ways in with M and D are together is 8. <no doubt here>

saxenashobhit wrote:I can select 3 people to make committee in 6C3 ways = 20 ways. Now these 3 people can join commitee 1 or committee 2. So total combinations should be - 2* 20 = 40 <this is what my question is - is this correct?>

Hi,
No..It should be just 20(not 40). 6C3 counts all possible triplets. So, if one triplet is fixed the other automatically goes into the other subcommittee(SC).
Let's say you have selected ABC are selected for SC1 .. DEF will go to SC2
While listing all triplets you will get DEF as well for SC!..So, ABC will go to SC2.
This means we are already taking care of both cases. So, we don't need to multiply by 2.

If you feel this is confusing, then it is better to follow the method think you are comfortable with(Mitch's method).