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board of directors

by GmatKiss » Fri May 18, 2012 11:19 am
Anthony and Michael sit on the six-member board of directors
for company 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?

· 20%
· 30%
· 40%
· 50%
· 60%
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by aneesh.kg » Fri May 18, 2012 2:03 pm
GmatKiss wrote:Anthony and Michael sit on the six-member board of directors
for company 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?

· 20%
· 30%
· 40%
· 50%
· 60%
Required probability/chances = (Number of subcommittees that include Michael and Anthony)/(Number of subcommittees that include both Michael)

Number of subcommittees that include Michael = (Number of ways of including Michael) AND (Number of ways of selecting two people from the remaining five people) = 1C1*5C2 = 10

Number of subcommittees that include both Michael and Anthony = (Number of ways of selecting Michael and Anthony) AND (Number of ways of selecting one more person from the remaining 4 people) = 1C1*4C1 = 4

Required probability/chances = 4/10 or 40%

[spoiler](C)[/spoiler] is correct
Last edited by aneesh.kg on Sat May 19, 2012 1:15 pm, edited 1 time in total.
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by Stuart@KaplanGMAT » Fri May 18, 2012 4:07 pm
GmatKiss wrote:Anthony and Michael sit on the six-member board of directors
for company 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?

· 20%
· 30%
· 40%
· 50%
· 60%
A great opportunity to use logic instead of math!

There are 5 different people with whom Michael can be matched. There are 2 spots available on Michael's team. Therefore, any one of the other 5 people will appear on 2/5 = 40% of Michael's teams. Pick C!
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by Anurag@Gurome » Fri May 18, 2012 10:45 pm
GmatKiss wrote:Anthony and Michael sit on the six-member board of directors
for company 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?

· 20%
· 30%
· 40%
· 50%
· 60%
The number of subcommittees having Michael is 5C2 = 10.
The number of subcommittees having both Michael and Anthony is 4C1 = 4.
So, required % is 4/10 * 100 = [spoiler]40%[/spoiler]

The correct answer is C.
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