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probability

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Re: probability

by cjb » Fri Feb 13, 2009 12:24 pm
yangliu0401 wrote:13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?
Is that the entire text of the question? My answer would be 1, since I'd put them on at the same time in the same place... Very unclear question, I think.
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Re: probability

by yangliu0401 » Fri Feb 13, 2009 4:18 pm
cjb wrote:
yangliu0401 wrote:13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?
Is that the entire text of the question? My answer would be 1, since I'd put them on at the same time in the same place... Very unclear question, I think.
It's a probability problem. My answer is 5/6, but the standard ans. is 2/3. Can someone explain in details?
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by cjb » Sat Feb 14, 2009 6:37 am
removed.
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by deepoe » Sat Feb 14, 2009 7:26 am
So you got a total of 7 keys.

That these 2 keys would be the first is 2/7 than the other key would be the second is 1/7. So 2/7 X 1/7 = 3/7 or am I wrong?
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by yangliu0401 » Sat Feb 14, 2009 1:06 pm
deepoe wrote:So you got a total of 7 keys.

That these 2 keys would be the first is 2/7 than the other key would be the second is 1/7. So 2/7 X 1/7 = 3/7 or am I wrong?
I've got the inspiration!!!
As the two keys were putting in a key CHAIN, there could be the following 3 situations:
1. they both in the very first;
2. they both in the very last;
3. one in the first and the other in the last.

As they are different two keys, so the total different situations are 3*2=6.
When the two keys were putting together, the situations could be 2*2=4

So the probability is 4/6=2/3 :o
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by deepoe » Sun Feb 15, 2009 2:03 am
yangliu0401 wrote:
deepoe wrote:So you got a total of 7 keys.

That these 2 keys would be the first is 2/7 than the other key would be the second is 1/7. So 2/7 X 1/7 = 3/7 or am I wrong?
I've got the inspiration!!!
As the two keys were putting in a key CHAIN, there could be the following 3 situations:
1. they both in the very first;
2. they both in the very last;
3. one in the first and the other in the last.

As they are different two keys, so the total different situations are 3*2=6.
When the two keys were putting together, the situations could be 2*2=4

So the probability is 4/6=2/3 :o
Why did you did this:

As they are different two keys, so the total different situations are 3*2=6 ?

The 2*2 is because of your reason 1 + 2?

Could you explain me:$? The rest of your explanation I understand :$
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by awesomeusername » Sun Feb 15, 2009 10:46 am
ABKKKKK
BAKKKKK
AKKKKKB
BKKKKKA
KKKKKAB
KKKKKBA

4/6 chance of them being together. This keyring thing is really tricky. I, at first, thought you could put the new keys wherever you wanted. But if you stop to think about how a keyring works, you see that it isn't possible.
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by deepoe » Mon Feb 16, 2009 2:29 am
tricky question yea.

My Uncle said it was like this:

ABKKKKK
KABKKKK
KKABKKK
KKKABKK
KKKKABK
KKKKKAB

6 options

en because the places could be switched between A and B = 2^2 = 4

So 4/6 = 2/3
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Re: probability

by x2suresh » Wed Feb 18, 2009 2:38 pm
yangliu0401 wrote:13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?
2!*5!/6! = 2/3
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Re: probability

by kamu » Thu Mar 12, 2009 12:08 am
yangliu0401 wrote:13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?
Simple if you could imagine.

Where can two new keys end up.

Both on one side.
Both on other side.
one on either side

2/3 probability of being together.
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by Reader » Sun Mar 15, 2009 10:43 am
1/3.

Once you put the first new key in, you now have 6 keys and 6 "gaps", only one you put the second new key in 2 of the 6 gaps will the new keys be together.

I think the OA is not logical or realistic at all, since they somehow think you can't put new keys in between old ones... hope real GMAT won't have it.
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by Nabil-Tun » Sun Mar 15, 2009 1:43 pm
13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?

It seems to me that we have to know, as well, that the five keys are not different; so that we can assume the five keys as KKKKK and not CDEFG and therefore the problem will be solved differently!
Please, if I misunderstood that, correct me!

Thanks,
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Re: probability

by rs2010 » Mon Mar 16, 2009 7:53 pm
x2suresh wrote:
yangliu0401 wrote:13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?
2!*5!/6! = 2/3
This will be 1/3 while OA is 2/3
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Re: probability

by Brent@GMATPrepNow » Fri Mar 20, 2009 4:12 pm
yangliu0401 wrote:13. Put two keys in a key chain which already has five keys. What is the probability that the two new keys will be together?
Here's my attempt via some graphics:
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