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by GMATGuruNY » Mon Oct 18, 2010 4:12 am
girishbtg wrote:Acc to me ,

as 2 of them ( A and M ) are fixed,
So ,

4C1 / 6 C 3 = 20 %
Only Michael is fixed.

Michael must be combined with 2 people from the 5 remaining choices.

Total possible pairs = 5C2 = 10.
Number of pairs that will include Anthony = 4 (since there are 4 other people who could be paired with Anthony).

P(pair with Anthony) = 4/10 = 40%.
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by girishbtg » Mon Oct 18, 2010 4:38 am
GMATGuruNY wrote:
girishbtg wrote:Acc to me ,

as 2 of them ( A and M ) are fixed,
So ,

4C1 / 6 C 3 = 20 %
Only Michael is fixed.

Michael must be combined with 2 people from the 5 remaining choices.

Total possible pairs = 5C2 = 10.
Number of pairs that will include Anthony = 4 (since there are 4 other people who could be paired with Anthony).

P(pair with Anthony) = 4/10 = 40%.
GMATGuruNY,

Right Now my mind is not working as am doing sums for last 4 yours... so am taking break for the day.. tmrw i will see yr solution...

You would be really great if you have made it... !
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by me10gmat800 » Mon Oct 18, 2010 4:48 am
Thanks GMATGuruNY for the explanation
GMATGuruNY wrote:
girishbtg wrote:Acc to me ,

Number of pairs that will include Anthony = 4 (since there are 4 other people who could be paired with Anthony).

P(pair with Anthony) = 4/10 = 40%.
I totally agree with the logic that the Anthony & Micheal pair will combine with 4 others to form a committee of 3.

However, won't the total number of possible sub-committees be formed in 6C3 ways?

And therefore the P will be 5C2/6C3 * 100 = 50%

Could you kindly explain if I am wrong?
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by GMATGuruNY » Mon Oct 18, 2010 6:49 am
me10gmat800 wrote:Thanks GMATGuruNY for the explanation
GMATGuruNY wrote:
girishbtg wrote:Acc to me ,

Number of pairs that will include Anthony = 4 (since there are 4 other people who could be paired with Anthony).

P(pair with Anthony) = 4/10 = 40%.
I totally agree with the logic that the Anthony & Micheal pair will combine with 4 others to form a committee of 3.

However, won't the total number of possible sub-committees be formed in 6C3 ways?

And therefore the P will be 5C2/6C3 * 100 = 50%

Could you kindly explain if I am wrong?
You're misreading the question, which asks:

Out of all the possible subcommittees that include Michael, what percent will also include Anthony?

Michael cannot be a member of 6C3 subcommittees, because not all of those subcommittees will include Michael. To form a 3-member subcommittee that includes Michael, he must be combined with 2 of the remaining 5 people. Thus, he can be a member only of 5C2 = 10 subcommittees.

For Anthony to be included as well, Anthony and Michael must be combined with 1 of the 4 remaining people, so they can serve together on 4C1 = 4 subcommittees.

P = good/total = 4/10 = 40%.

Does this help?
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