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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] New post 16 Jun 2006, 21:37
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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 into 4 envelopes at random, what is the probability that only 1 letter will be put into envelope with its correct address?

a. 1/24

b. 1/8

c. 1/4

d. 1/3

e. 3/8

Last edited by mrmikec on 17 Jun 2006, 10:13, edited 1 time in total.
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 [#permalink] New post 16 Jun 2006, 22:26
D.

Total no. of ways of posting = 4!

Now, if one letter is in the correct envelope, then 2nd letter can go into 2 of the 3 remaining envelopes (as the 3rd is the correct envelope), and so on... and there are 4 ways in which this can be done. So, we get

No. of ways only one letter is correctly posted = 4 x (2 x 1 x 1)

Probability = 1/3
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 [#permalink] New post 17 Jun 2006, 10:22
paddyboy wrote:
D.

Total no. of ways of posting = 4!

Now, if one letter is in the correct envelope, then 2nd letter can go into 2 of the 3 remaining envelopes (as the 3rd is the correct envelope), and so on... and there are 4 ways in which this can be done. So, we get

No. of ways only one letter is correctly posted = 4 x (2 x 1 x 1)

Probability = 1/3


i'm sorry i don't understand your answer.

I'm getting 1/4 because say

the first letter.. 3/4 chance it goes into the wrong envelope. 4 letter/ 3 incorrect envelopes

the second letter.. 2/3

the third letter... 1/2

3/4 * 2/3 * 1/2 = 6/24 = 1/4
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 [#permalink] New post 17 Jun 2006, 10:37
Assume one letter is in the correct envelope.
Probability of next letter in incorrect envelope is 2/3
Probability of one out of 2 remaining letters is 1/2
Probability of last is 1

Probability of first letter to be in correct envelope is 1/4. And this can happen in 4 ways (four letters).

So probability = 4*(1/4)*(2/3)*(1/2)*(1/1) = 1/3. (D) Is it right or wrong? :) Too much suspense :lol:
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 [#permalink] New post 17 Jun 2006, 10:44
paddyboy wrote:
Assume one letter is in the correct envelope.
Probability of next letter in incorrect envelope is 2/3
Probability of one out of 2 remaining letters is 1/2
Probability of last is 1

Probability of first letter to be in correct envelope is 1/4. And this can happen in 4 ways (four letters).

So probability = 4*(1/4)*(2/3)*(1/2)*(1/1) = 1/3. (D) Is it right or wrong? :) Too much suspense :lol:


This is from GMATPrep so I don't have the OA written down. I will look got it though
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 [#permalink] New post 17 Jun 2006, 19:52
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