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probability question

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by jamesk486 » Fri Aug 15, 2008 2:40 pm
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%


Can someone explain how to do this problem?
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by pepeprepa » Fri Aug 15, 2008 2:58 pm
Total possibilities: 6C3
Possibilities with Anthony and Michael in the same subcommitee is: 4

4/20=1/5=20%

Hope it's ok
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by arorag » Sat Aug 16, 2008 9:34 am
My random guess is 50 %
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by sudhir3127 » Sat Aug 16, 2008 9:41 am
IMO C..

6 members and two committees with 3 each.

Michael will be on one of the committees out of 6
So, Anthony has 5 positions to be in. He can be in any of the 5 with equal probability and 2 out of 5 will share it with Anthony

So 2/5 = 40%

do let us know the OA..
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by pepeprepa » Sat Aug 16, 2008 9:54 am
RC problem in my first post.

There are 5C2 subcommitees with Michael in.
4 can also have Anthony.

So 4/10 or 40% yep
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by 4meonly » Sat Aug 16, 2008 10:57 am
pepeprepa wrote:RC problem in my first post.

There are 5C2 subcommitees with Michael in.
4 can also have Anthony.

So 4/10 or 40% yep

Why 4 can also have Anthony?
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by pepeprepa » Sat Aug 16, 2008 11:07 am
Because in my view, I am not sure about the answer I don't want to mislead you, we have:
M A 1 2 3 4 people

MA1
MA2
MA3
MA4

4 groups in which M and A are together.
These 4 groups are among the 5C2 ones.
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by hengirl03 » Sat Aug 16, 2008 11:15 am
What's the OA?
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by 4meonly » Sat Aug 16, 2008 12:01 pm
Ye, I-ve got it!! ))
pepeprepa, thank you!
4 from 20 is 20%
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by AlaaQasem » Sat Aug 16, 2008 5:21 pm
I think the answer is 20% as well. Here is how I solved it. I'm not sure of the answer, though.

What we need to find is:
(The probability of having Anthony in sub-committee 1 AND the probability of having Michael in sub-committee 1)
OR
(The probability of having Anthony in sub-committee 2 AND the probability of having Michael in sub-committee 2)

Mathematically, this will translate into:
The probability of having Anthony in sub-committee 1 = count of individual to select/count of available individuals to select from = 1/6
The probability of having Michael in sub-committee 1 = count of individual to select/count of available individuals to select from exculding Anthony = 1/5
The probability of having Michael in sub-committee 2 = count of individual to select/count of available individuals to select from exculding the three individuals already assigned to sub-committtee 1 = 1/3
The probability of having Michael in sub-committee 2 = count of individual to select/count of available individuals to select from exculding the three individuals already assigned to sub-committtee 1 and Anthony = 1/2

Now substitute the numbers above:
(1/6)*(1/5)+(1/3)*(1/2) = (1/30) + (1/6) = (1+5)/30 = 6/30 = 1/5 = 20%
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by sudhir3127 » Sat Aug 16, 2008 7:59 pm
Hi jamesk486..i would be nice if u can post the OA. I request Ian/ Stuart to pitch in and clear the confusion. For me its still 40%.
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by pepeprepa » Sun Aug 17, 2008 1:21 am
4meonly
What I did is:
- to count the total number of subcommitte groups we can make with Michael in. If you want we can write it like " M _ _ ". You can see that there are two available seats and there are 5 guys who can come there. So that is why I think the total possibilities are 5C2.
5C2= 5!/(2!3!)=(5*4)/2=10
- to count the number of groups of Michael in which there is also Anthony: 4

Finally, 4/10 or 40%
Hope it is ok
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by 4meonly » Sun Aug 17, 2008 4:11 am
pepeprepa wrote:4meonly
What I did is:
- to count the total number of subcommitte groups we can make with Michael in. If you want we can write it like " M _ _ ". You can see that there are two available seats and there are 5 guys who can come there. So that is why I think the total possibilities are 5C2.
5C2= 5!/(2!3!)=(5*4)/2=10
- to count the number of groups of Michael in which there is also Anthony: 4

Finally, 4/10 or 40%
Hope it is ok
5C2? I've made so:
All possibilities to make group of 3 from 6 is 6C3 = 6!/(3!3!)=20
We can have only 4 grous with M and A, so 4/20=20%

I cant find mistake in my reasoning. :(
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by pepeprepa » Sun Aug 17, 2008 4:52 am
For me, with 6C3 you count the total number of possibilities
With 5C2, you count all the commitees in which you have Michael.
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by pranavc » Sun Aug 17, 2008 3:24 pm
IMO A.
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