Manhattan Probability tougie

girishbtg wrote:Acc to me ,

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

4C1 / 6 C 3 = 20 %

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 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?

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?