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Letters (Probability)

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Joined: 16 Jul 2009
Posts: 1481

Kudos [?]: 1527 [0], given: 2

Schools: CBS
WE 1: 4 years (Consulting)
Letters (Probability) [#permalink]

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New post 25 Jul 2009, 07:32
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Tanya prepared 4 different letter to be senmt 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?
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Kudos [?]: 1527 [0], given: 2

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Joined: 18 Jul 2009
Posts: 167

Kudos [?]: 116 [0], given: 37

Location: India
Schools: South Asian B-schools
Re: Letters (Probability) [#permalink]

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New post 25 Jul 2009, 10:52
let us consider there are four letters (L1,L2,L3,L4) with corresponding right addressed envelopes (e1,e2,e3,e4)

Now if we assume atleast one correct

Consider => L1-e1 is right => then L2 has 2 possible combinations such as e3 or e4 (we consider L2-e3); it cannot take e2 bcoz then there will be a combination with 2 letters in right envelop. Now the obvious choice for L3 to take e4 and not e3 (same logic for L2) and thus L4 is paired bydefault with e2.....Now plz refer below sequence with 2 possibilities

Hence => [L1-e1 | L2-e3 | L3-e4 | L4-e2] OR [L1-e1 | L2-e4 | L3-e2 | L4-e3]

Now this sequence is only for L1-e1 to be the only right and all others are wrong; similarly we have 4 pairs such as (L1-e1 | L2-e2 | L3-e3 | L4-e4 ) with 2 possibilities for each

Hence Ans is 4 x 2 = 8 => 8/24 => 1/3...(plz let me know if i am right)

Plz consider for Kudos if right....(trying to access tests :wink: )
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Last edited by bhushan252 on 25 Jul 2009, 11:09, edited 1 time in total.

Kudos [?]: 116 [0], given: 37

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Kudos [?]: 96 [1], given: 13

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Re: Letters (Probability) [#permalink]

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New post 25 Jul 2009, 11:03
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I got 1/3

Usually in problems like this, it is easy to make mistakes. But here is a first attempt.

Four spaces, which are the four envelopes:
_ _ _ _
Number of total possibilities P4 = 24

Now, number of possibilities that ONLY the letter A is in its respective envelope, the first space:
A _ _ _
A C D B - 1 possibility
A D B C - 1 possibility

Just as there are only 2 possibilities for ONLY the letter A is in its respective envelope, so there are only 2 possibilities for ONLY each one of the others letters be in its respective envelope.

Summing we have 8 possibilities.

8/24 = 1/3.

I hope this is right ;) If it is, I accept a kudo, so I can access the GMATClub tests !!!

Kudos [?]: 96 [1], given: 13

Re: Letters (Probability)   [#permalink] 25 Jul 2009, 11:03
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Letters (Probability)

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