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Anthony and Michael

by sanju09 » Sat Feb 21, 2009 2:22 am
Anthony and Michael sit on the six member board of directors for compnay X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%
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by sureshbala » Sat Feb 21, 2009 2:39 am
The total subcommittees that will contain Michael is 5C2 = 10.

Out of these total subcommittees that will contain Anthony also = 4.

So the percentage has to be 4/10 = 40%
Last edited by sureshbala on Sat Feb 21, 2009 3:40 am, edited 1 time in total.
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by sanju09 » Sat Feb 21, 2009 3:10 am
sureshbala wrote:The total number of ways of dividing 6 people into 2 groups of 3 members each is 6C3 = 20.

Now we want Michael and Anthony to be in the same group. So the other person of this group could be anyone from the remaining 4. So favourable cases will be 4.

Hence the percentage is 4/20 = 20%
:) If we take M and A for Michael and Anthony respectively, and X for any one member from the remaining four; then given that M is already in, second and third entries could be either A and X, or X and A. It is a conditional probability sum.
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by sureshbala » Sat Feb 21, 2009 3:38 am
Got it...I must have read the question with a bit of patience....

The total subcommittees that will contain Michael is 5C2 = 10.

Out of these total subcommittees that will contain Anthony also = 4.

So the percentage has to be 4/10 = 40%
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