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2 three-person subgroups

by sanju09 » Wed Mar 02, 2011 5:05 am
Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
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by Night reader » Wed Mar 02, 2011 5:38 am
3(Bob)*2(Allen)*(some one else) OUT OF 6C3 --> 6/20=3/10=0.3 (30%)

IOM B
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
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by manpsingh87 » Wed Mar 02, 2011 9:45 am
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
IMO A..!!!
Total no. of ways of forming sub groups; 6!/3!*(2!)^3 = 15

Total no. of favorable cases; 3

hence required probability or percentage is 3/15 = 1/5 = 20% [spoiler]hence A..!!![/spoiler]

Please post correct answer...!!!
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by GMATGuruNY » Wed Mar 02, 2011 10:03 am
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
Number of 3-member groups that include Bob:
Bob can be combined with 5C2 = 10 other pairs.

Number of 3-member groups that include both Bob and Allen:
Bob and Allen can be combined with 4C1 = 4 other people.

Out of all the groups with Bob, 4/10 = 40% will also include Allen.

The correct answer is C.
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by kevincanspain » Wed Mar 02, 2011 2:33 pm
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
In other words, what is the probability that Allen will be one of the two people who accompany Bob? Since apart from Bob, there are 5 people, the required probability is 2/5
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by anu009 » Sat Mar 05, 2011 8:58 pm
Hello Instructors

The probabilty & combinatin questions are really confusing for me.Each time there is a different way to solve the problem to get to the correct solution.

I was trying to do this with basic concepts of Single source order doesnt matter n!/(n-r)!*r! which was giving 6C3 = 15 and the favourable cases = 3. SO favourable/total probabilyt = 3/15 * 100 = 20% but the answer is different.

When i saw your solution i found it rt but i am not sure where was i wrong.Can you help me with this and so that the general rule of solving the questions works everywhere.

Thank you.
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by Anurag@Gurome » Sat Mar 05, 2011 9:22 pm
anu009 wrote:Hello Instructors

The probabilty & combinatin questions are really confusing for me.Each time there is a different way to solve the problem to get to the correct solution.

I was trying to do this with basic concepts of Single source order doesnt matter n!/(n-r)!*r! which was giving 6C3 = 15 and the favourable cases = 3. SO favourable/total probabilyt = 3/15 * 100 = 20% but the answer is different.

When i saw your solution i found it rt but i am not sure where was i wrong.Can you help me with this and so that the general rule of solving the questions works everywhere.

Thank you.
Probability is (no. of favorable cases)/(total no. of all possible cases).

Here, what you have taken as "total no. of all possible cases" is "total ways of selecting a group of 3 from among 6 people".
Read the question again.
It is "what percent of all possible subgroups that include Bob".
So, we need to estimate as "total no. of all possible cases", the number of groups of 3, that always have Bob.
This can be done in 5C2 or 10 ways. ( this is because leaving Bob, we have to select other 2 from remaining 5).
Also "no. of favorable cases" is number of ways we can select groups of 3 which have Bob and Allen both.
This can be done in 4C1 or 4 ways (this is because we have to select 1 more person from remaining 4 apart from Bob and Allen).
So required prob. is 4/10 = 40%.
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by anu009 » Sat Mar 05, 2011 9:31 pm
Got it. Thank You so much.
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by vineeshp » Sat Mar 05, 2011 9:45 pm
kevincanspain wrote:
In other words, what is the probability that Allen will be one of the two people who accompany Bob? Since apart from Bob, there are 5 people, the required probability is 2/5
Kevin,
Can you explain your solution a bit more?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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