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Tanya prepared 4 different letters to 4 different addresses

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4 letters 4 addresses probablity problem [#permalink] New post 17 Sep 2012, 17:09
Max 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 only 1 letter will be put into the envelope with its correct address? answer is 1/3 but very confused.
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Re: 4 letters 4 addresses probablity problem [#permalink] New post 17 Sep 2012, 23:39
clearmountain wrote:
Max 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 only 1 letter will be put into the envelope with its correct address? answer is 1/3 but very confused.


Yes, it is a little tricky. That is why I have discussed it in detail in this post:
http://www.veritasprep.com/blog/2011/12 ... envelopes/
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Re: 4 letters 4 addresses probablity problem [#permalink] New post 18 Sep 2012, 00:04
clearmountain wrote:
Max 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 only 1 letter will be put into the envelope with its correct address? answer is 1/3 but very confused.


4 is not a big number, so you can deal with the computations directly.

The total number of possibilities to place the 4 letters in 4 envelopes is 4!.

Any one of the 4 letters can be placed correctly, so 4 possibilities.
After 1 letter is placed in its correct envelope, think of how many ways are there to place the remaining 3 letters such that neither one is in its correct envelope.
3! is the number of possibilities to arrange the three letters, say A,B,C in their envelopes.
Out of these, one possibility when each letter is in its correct envelope - 1
Then, there are 3 more possibilities when one letter is correctly placed and the other two are switched between them - 3
Therefore, a total of 3! - 4 possibilities to have exactly one letter placed correctly.

Required probability is 4(3! - 4)/4! = 4*2/(2*3*4) = 1/3.
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Last edited by EvaJager on 18 Sep 2012, 00:59, edited 1 time in total.
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Re: 4 letters 4 addresses probablity problem [#permalink] New post 18 Sep 2012, 00:42
clearmountain wrote:
Max 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 only 1 letter will be put into the envelope with its correct address? answer is 1/3 but very confused.


Merging similar topics.
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Re: 4 letters 4 addresses probablity problem   [#permalink] 18 Sep 2012, 00:42
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