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Manhattan cat1 ---Probability --committee of three ppl

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Manhattan cat1 ---Probability --committee of three ppl

by prachich1987 » Sat Dec 25, 2010 4:11 am
Hi,

I know the below question had already been placed on the forums.
But I just wanted to ask if instead of asking "what percent of all the possible subcommittees that include Michael also include Anthony" ,if the question asks "what percent of all the possible subcommittees will have Michael & Anthony together "
Will the answer for later be 50??Plz help

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 anshumishra » Sat Dec 25, 2010 7:09 am
prachich1987 wrote:Hi,

I know the below question had already been placed on the forums.
But I just wanted to ask if instead of asking "what percent of all the possible subcommittees that include Michael also include Anthony" ,if the question asks "what percent of all the possible subcommittees will have Michael & Anthony together "
Will the answer for later be 50??Plz help

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%
It would be same (40%), not solving as you know the answer already.
what percent of all the possible subcommittees will have Michael & Anthony together - basically means the same that the subcommittee contains both of them.
If you have to place them on chairs such that Michael & Anthony have to be side by side then the arrangement scenario comes to picture.
Thanks
Anshu

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by prachich1987 » Sat Dec 25, 2010 8:36 am
anshumishra wrote:
prachich1987 wrote:Hi,

I know the below question had already been placed on the forums.
But I just wanted to ask if instead of asking "what percent of all the possible subcommittees that include Michael also include Anthony" ,if the question asks "what percent of all the possible subcommittees will have Michael & Anthony together "
Will the answer for later be 50??Plz help

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%
It would be same (40%), not solving as you know the answer already.
what percent of all the possible subcommittees will have Michael & Anthony together - basically means the same that the subcommittee contains both of them.
If you have to place them on chairs such that Michael & Anthony have to be side by side then the arrangement scenario comes to picture.
Thanks for the above
But I think the two probabilities should be different.
I mean,

"what percent of all the possible subcommittees that include Michael also include Anthony"----Here we are calculating the % with the base as "no. of teams having Michael"..Hence the answer would be 40.

"what percent of all the possible subcommittees will have Michael & Anthony together "----Here we are calculating the % with the base as " total no. of teams..& I think here the answer should be 50%
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by beat_gmat_09 » Sat Dec 25, 2010 7:37 pm
prachich1987 wrote: Thanks for the above
But I think the two probabilities should be different.
I mean,

"what percent of all the possible subcommittees that include Michael also include Anthony"----Here we are calculating the % with the base as "no. of teams having Michael"..Hence the answer would be 40.

"what percent of all the possible subcommittees will have Michael & Anthony together "----Here we are calculating the % with the base as " total no. of teams..& I think here the answer should be 50%
Hi,
Start by selecting one member, say anthony, probability of putting anthony on committee A is - 1/2 as there are two committees.
Probability of putting michael on the same committee A as anthony is = 2/5
Combined = 1/2*2/5 = 1/5
Now take the other case, probability of putting anthony on committee B is = 1/2
same way, Probability of putting michael on the same committee B as anthony is = 2/5
Combined = 1/2*2/5 = 1/5
This is OR situation, required probab = 1/5 + 1/5 = 2/5 which is 40%
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by prachich1987 » Sun Dec 26, 2010 2:34 am
beat_gmat_09 wrote:
prachich1987 wrote: Thanks for the above
But I think the two probabilities should be different.
I mean,

"what percent of all the possible subcommittees that include Michael also include Anthony"----Here we are calculating the % with the base as "no. of teams having Michael"..Hence the answer would be 40.

"what percent of all the possible subcommittees will have Michael & Anthony together "----Here we are calculating the % with the base as " total no. of teams..& I think here the answer should be 50%
Hi,
Start by selecting one member, say anthony, probability of putting anthony on committee A is - 1/2 as there are two committees.
Probability of putting michael on the same committee A as anthony is = 2/5
Combined = 1/2*2/5 = 1/5
Now take the other case, probability of putting anthony on committee B is = 1/2
same way, Probability of putting michael on the same committee B as anthony is = 2/5
Combined = 1/2*2/5 = 1/5
This is OR situation, required probab = 1/5 + 1/5 = 2/5 which is 40%
Thanks for above.
Will you plz help me by answering below question.

Q: There are total six people & we want to form two committees of three people each.What are the no of ways of doing this?
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