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# Tanya prepared 4 different letters to be sent to 4 different

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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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10 Dec 2006, 08:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that exactly 2 letters will be put into 2 envelopes with its correct address?
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Impossible is nothing

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10 Dec 2006, 08:36
Is it 1/4?

Chance of 1st letter being correct = 1/4
Chance of 2nd letter being correct = 1/3
Chance of 3rd letter being incorrect = 1/2
Chance of 4th letter being incorrect =1

1/4 * 1/3 * 1/2 * 1 = 1/24

No of ways of choosing 2 letters from 4 = 4C2 = 6

6*1/24 = 1/4

Where is the question from?

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10 Dec 2006, 11:21
I also got 1/4

The original question is from GMATprep but I changed it a bit to spice things up. Wanted to check if I fully understood the concept
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10 Dec 2006, 12:32
is 1/4 the correct answer though ? - the question said any two letters need to be correctly addressed not the first two ....

this is what I got ...
number of ways 4 letters could be arranged in 4 envelopes = !4 = 24

let any of the two letters be correctlu put into the envelope ,
Now the number of ways the rest two letters could be put = !2 = 2

So, the required probability is 2/24 = 1/12

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10 Dec 2006, 13:17
I tabulated the possibilities and for any two to be correct the answer is 1/4

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10 Dec 2006, 15:15

The total number of possibilities : 24 (4p4)
4 envelopes and 4 letters

The possibility that 1 and 2 envelope has the right letters = 2P2 = 2

There are a total of 7 such possibilities: 12, 13, 14, 23, 24, 34, 14 : 14

The probability : 14/24 = 7/12

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10 Dec 2006, 16:19
There is no OA cos adaptation of GMAT prep question that asks same thing but for 1 letter in correct position. I've drawn a table of all posibilities - correct answer is 1/4

I've labelled envelopes, 1,2,3,4 with correct letter being A=1, B=2, C=3, D=4
Attachments

ProbTable.doc [54 KiB]

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10 Dec 2006, 17:28
I agree... That is a concrete way to prove something

My mistake..

i took in to the choices when all 4 are correct into the count..

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10 Dec 2006, 17:49
I still wonder about the method:

Maybe it should be total no. of ways = 4! = 24
No of unique ways to choose letters = 4C2 = 6

6/24=1/4?

Maybe one of the math gurus could help -- Fig?

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10 Dec 2006, 21:44
No I agree it should be 1/4 ...

4C2/ 4! = 6/24 = 1/4

The number of ways any two letters can be correctly put out of total 4 envelopes should be 4C2 ...

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11 Dec 2006, 09:44
Thanks. I'm just wondering if the original way I calculated it (1/4*1/3*1/2*1*4C2=1/4) is a sound method.

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11 Dec 2006, 12:03
Getting 1 / 4

Total number of cases - 4! = 24
Combination of 2 letters = 4 C 2 = 6
Putting 2 letters in correct address can happen only in one way = 6 * 1 = 6

So 6/24 = 1/4

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11 Dec 2006, 12:14
anindyat wrote:
Getting 1 / 4

Total number of cases - 4! = 24
Combination of 2 letters = 4 C 2 = 6
Putting 2 letters in correct address can happen only in one way = 6 * 1 = 6

So 6/24 = 1/4

agree with anidya it must be 1/4

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11 Dec 2006, 12:14
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