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probability question from Gmatprep CAT

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probability question from Gmatprep CAT

by thegmatbeater » Sun Jul 20, 2008 11:34 am
Jane prepared 4 different letters to 4 different addresses. Fer each letter she prepared an envelop with its correct address. If 4 letters are to be put into 4 envelopes at random, what is the probability that only one letter will be put into envelop with correct address?

a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8
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Re: probability question from Gmatprep CAT

by parallel_chase » Sun Jul 20, 2008 12:01 pm
thegmatbeater wrote:Jane prepared 4 different letters to 4 different addresses. Fer each letter she prepared an envelop with its correct address. If 4 letters are to be put into 4 envelopes at random, what is the probability that only one letter will be put into envelop with correct address?

a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8
Is the answer 1/8?
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by thegmatbeater » Sun Jul 20, 2008 12:18 pm
no, it is 1/3
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by pepeprepa » Sun Jul 20, 2008 1:13 pm
What I do is to write it with letters.
For example, "ABCD" means letter A is in the right envelop (1st place), B 2nd place is right, C 3rd place is right, D 4th place is right. So ABCD is the case when 4 letters are in the 4 good envelops.
And after I write all possibilities for A at the first place.
ABCD
ABDC
ACBD
ACDB (*)
ADCB
ADBC (*)
And I look at the number of combinations which only have 1 letter at the right place: 2
We have 4 letters so 4x2=8 total good possibilities
And total of 24 possibilities
8/24=1/3

But I don't like that, someone has a calculate explanation?
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by parallel_chase » Sun Jul 20, 2008 1:44 pm
pepeprepa wrote:What I do is to write it with letters.
For example, "ABCD" means letter A is in the right envelop (1st place), B 2nd place is right, C 3rd place is right, D 4th place is right. So ABCD is the case when 4 letters are in the 4 good envelops.
And after I write all possibilities for A at the first place.
ABCD
ABDC
ACBD
ACDB (*)
ADCB
ADBC (*)
And I look at the number of combinations which only have 1 letter at the right place: 2
We have 4 letters so 4x2=8 total good possibilities
And total of 24 possibilities
8/24=1/3

But I don't like that, someone has a calculate explanation?
I did it in a different manner, took me about 15 mins though.

The probability of A letter going to A envelop and B, C, and D in wrong

A going into right envelop is 1/4
B going into wrong envelop is 2/3 ( 1-1/3)
C going into wrong envelop is 1/2
D automatically will go into the wrong envelop 1

1/4*2/3*1/2*1 = 1/12 (multiplication because everything occurs at once)

Similarly the same procedure will follow for B, C & D and the result will also be the same

1/12+1/12+1/12+1/12 = 4/12 = 1/3.

I am not too sure about the method but it seems correct. Let me know if you guys think otherwise.
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by [email protected] » Sat Aug 02, 2008 12:57 am
Does anyone know a quicker way of doing this - that follows some sort of strategy?

Thanks in advance!
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by egybs » Sat Aug 02, 2008 1:32 am
This is a little tougher than some of your usual probability questions... but fairly simple when you break it down.

We're looking for 1 letter to be placed in the right envelope. So there are four envelopes where the correct letter to be placed.

So let's start with the placement of the first envelope (the correctly placed one): Randomly choosing a letter gives you a 1/4 chance of getting the right envelope.

2nd envelop, we're looking for a mismatched letter... So this is going to be 2/3 because there are 2 incorrectly addressed envelopes out of 3.

For the third envelop, there is a 1/2 chance of picking the mismatched letter. Notice that whichever you pick here, if it's incorrectly matched, the final envelop will also be incorrectly matched... so there's no need to go further.

So you get:

(1/4)*(2/3)*(1/2) * 4 = 1/3 (the four comes from the fact that there are four envelopes where you could match one letter correctly).


Notice that you end up not needing to worry about the 1/4 and the 4 as they cancel each other out. All you really need to think about is the 2nd and 3rd placements.

Once you see how this works, it's a really quick problem.
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by [email protected] » Sat Aug 02, 2008 2:23 am
Great thanks for the explanation.
I am still a little unsure why you multiply by 4 at the end?
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by Ian Stewart » Sat Aug 02, 2008 2:29 am
Pretty much the same as methods above, but a bit simpler (there's no need to label the letters or envelopes):

One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3.
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by egybs » Sat Aug 02, 2008 2:30 am
Assume there are four envelopes, A,B,C and D and four letters, 1,2,3 and 4.

A goes with 1, B goes with 2, etc.

Given that we want only one to be in the proper envelope this could happen in any one of four ways:

1 being in A, 2 being in B, 3 being in C, or 4 being in D.

So if the question were to be something like: "what's the probability that the only correctly placed letter is #1?" Then you'd say:
1/4 chance that I pick 1 in A, a 2/3 chance that I don't pick 2 in B, and a 1/2 chance that I don't pick 3 in C and 4 in D. All together that would be 1/12.

But the question isn't limiting it exclusively to 1 being in A... any one of the letters could be put in their respective envelopes... so you multiply by 4.

Hope this helps.
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by egybs » Sat Aug 02, 2008 2:31 am
Oh Ian, I already said that! :-)

"Notice that you end up not needing to worry about the 1/4 and the 4 as they cancel each other out. All you really need to think about is the 2nd and 3rd placements. "
Ian Stewart wrote:Pretty much the same as methods above, but a bit simpler (there's no need to label the letters or envelopes):

One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3.
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by sudhir3127 » Sat Aug 02, 2008 6:07 am
Even i got 1/3

heres how i did it ...

only 1 letter will be put into the envelope with its correct address 1 is right , other 3 are not.
total ways = 4 !

consider any one letter getting into right envelope , this can be done in 1 way, for next letter to be into wrong envelope we have 2 choices, for the next letter to be into wrong envelope we have 1 choice and for the 4th letter we have again 1 choice to put it into the wrong envelope.

and we have such 4 letters so 4(1*2*1*1) / 4! = 1/3
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