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# Letters (Probability)

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VP
Joined: 16 Jul 2009
Posts: 1248
Schools: CBS
WE 1: 4 years (Consulting)

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25 Jul 2009, 08:32
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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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Manager
Joined: 18 Jul 2009
Posts: 166
Location: India
Schools: South Asian B-schools

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Updated on: 25 Jul 2009, 12:09
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 )
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Originally posted by bhushan252 on 25 Jul 2009, 11:52.
Last edited by bhushan252 on 25 Jul 2009, 12:09, edited 1 time in total.
Manager
Joined: 03 Jul 2009
Posts: 103
Location: Brazil

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25 Jul 2009, 12:03
1
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 !!!

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Letters (Probability)   [#permalink] 25 Jul 2009, 12:03
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# Letters (Probability)

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