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Tanya prepared 4 different letters to 4 different addresses

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Tanya prepared 4 different letters to 4 different addresses [#permalink]  31 Dec 2009, 20:22
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Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jul 2012, 01:44, edited 2 times in total.
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Re: Combinations/Permutations [#permalink]  19 Oct 2010, 11:07
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dk94588 wrote:
Hello, this was on GMATprep, and I have had problems with this type of question before, but maybe you could help me solve it.

Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8

Let's call the envelopes E1, E2, E3 and E4.

P(only E1 gets the correct letter):

P(E1 gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(E2 gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong)
P(E3 gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(E4 gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in either E2 or E3)

Since we need all of these events to happen, we multiply the fractions:

1/4 * 2/3 * 1/2 * 1/1 = 1/12.

Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.

[Reveal] Spoiler:
D
.
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Re: Tanya prepared [#permalink]  19 Oct 2010, 10:45
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1/3

Only 1 letter finds the correct envelope --- that can happen in 4 cases when each of the 4 letters fills into the correct envelope

So, 1st envelope: can be filled in 4 ways (i.e., 4 possible correct letters)

The other 3 go astray:

2nd envelope: can be filled in 2 ways (2 wrong letters)
3rd envelope: can be filled in 1 way (1 wrong letter)
4th and final envelope: can be filled in 1 way (1 wrong letter)

So number of desired events = 4*2*1*1 = 8

Total ways to place 4 letters in 4 envelopes = 4*3*2*1 = 24

Prob = Desired / Total = 8/24 = 1/3
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Tanya prepared 4 different letters to 4 different addresses [#permalink]  31 Dec 2009, 20:45
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dk94588 wrote:
Hello, this was on GMATprep, and I have had problems with this type of question before, but maybe you could help me solve it.

Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8

Total # of ways of assigning 4 letters to 4 envelopes is $$4!=24$$.

Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct).

ABCD(envelopes)
ACDB(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

$$P(C=1)=\frac{8}{24}=\frac{1}{3}$$

All other possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html

Hope it's clear.
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Re: Combinations/Permutations [#permalink]  01 Apr 2012, 22:59
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Bunuel wrote:
dk94588 wrote:
Hello, this was on GMATprep, and I have had problems with this type of question before, but maybe you could help me solve it.

Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8

Sorry if I am replying to a very old post, But I know a formula which can be used for the above problem.

The above problem comes under a category called DERANGEMENT: Where you basically asked to match objects with WRONG pairs. Like putting a letter into incorrect envolope

Formula is pretty straight forward:

Lets consider this question, where you are asked to put only 1 letter in the correct envelope that implies that you need to put the other THREE letters in incorrect envelopes:

d(n) = $$n!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}.......\frac{(-1)^n}{n!})$$

D(n) = $$\frac{d(n)}{n!}$$

in this case we need to calculate for n=3

d(3) = $$3! * (\frac{1}{2!}-\frac{1}{3!})$$ = 2
D(3) = $$\frac{d(3)}{3!}$$ = $$\frac{1}{3}$$

Plese note that the only condition here is n >= 2, As you know that probality of putting only 1 letter in incorrect envelope is ZERO as it is not possible situation
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  31 Jul 2012, 10:09
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My 2 cents:

For 1 letter to go in right envelope - probability = 1/4
For 2nd letter to go in wrong envelope - probability = 2/3
For 3rd letter to go in wrong envelope - probability = 1/2
For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done:
C - Letter going into correct envelope
W - Letter going into wrong envelope

No#1. C - W - W - W
No#2. W - C - W - W
No#3. W - W - C - W
No#4. W - W - W - C

There are 4 ways, it can be done

So, probability of 1 letter going in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  04 Aug 2012, 12:38
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raghupara wrote:
Yes, the total # of possible placements for the three letters is 3! = 6 in general, but in this case when we deal with the probability of these letters not being mapped to the right address, we have to take out the right one and so it becomes 2! = 2/letter.
Also, I did not have to compute the same again for other letters, meaning, P(A alone mapped to the right address, but the rest not) = P(B alone mapped to right address while others are not)........ = 2*2*2/4! = 1/3?

(1) Using combinatorics:
To place one letter correctly (say A) - 1 possibility
The second letter to misplace it - 2 possibilities
The third letter to misplace it - 1 possibility
The fourth letter will be surely misplaced - just 1 possibility left
Consider the above scenario 4 times - 4 different possibilities for the correctly placed letter (A, B, C, or D can be placed correctly)
Therefore, the probability for exactly one letter placed correctly is 4 * 2 * 1 * 1/4! = 8/24 = 1/3

(2) Using probabilities:
To place one letter correctly (say A) - 1/4
The second letter to misplace it - 2/3
The third letter to misplace it - 1/2
The fourth letter will be surely misplaced - 1/1
So, the probability of a certain letter to be placed correctly and all the others to be misplaced is
(1/4)*(2/3)*(1/2) *(1/1) = 1/12
Again, consider the above scenario 4 times - it gives 4*(1/12) = 1/3

In your answer there are no probabilities (probabilities are expressed as fractions, between 0 and 1 inclusive).
Using the fundamental formula for probabilities (the number of desired outcomes / the total number of possible outcomes) again, your explanation isn't correct.
1*2*2*2 doesn't match any suitable scenario for placing exactly one later correctly and misplace all the other three.
Pay attention to getting the correct answer by using the correct reasoning.
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Re: Combinations/Permutations [#permalink]  18 Oct 2011, 20:36
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A more cumbersome but relatively straight fwd approach for probability-sick people like me

General counting method should work fine here (since we're dealing with only 4 envelopes)

ABCD
ABDC
ACBD
*ACDB

BACD
BDAC
*BDCA
BCDA

*CABD
CDAB
CDBA
*CBDA

*DACB
DABC
DBAC
*DBCA
DCBA
DCAB

Only arrangements with * marked against them have 1 letter correctly places in its corresponding address, hence probability = 8/24 = 1/3. Hence (D).
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  04 Aug 2012, 05:10
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Bunuel wrote:
BN1989 wrote:
What if two letters are placed correctly, what's the probability of that? 1/6?

@pkonduri what's the general forumula for finding the derangements when none is placed correctly?

Check this for all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html

Hope it helps.

Hi,

Is my understanding correct?

Total # of ways - 4! = 24.

Desired:
A-Mapped to the only correct address----------x 1 way only
B-Mapped to other two incorrect addresses - x 2 ways
C-Mapped to other two incorrect addresses - x 2 ways
D-Mapped to other two incorrect addresses - x 2 ways

Therefore, 1*2*2*2/24 = 1/3.?

Thanks.
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  09 Aug 2012, 16:26
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GODSPEED wrote:
My 2 cents:

For 1 letter to go in right envelope - probability = 1/4
For 2nd letter to go in wrong envelope - probability = 2/3
For 3rd letter to go in wrong envelope - probability = 1/2
For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done:
C - Letter going into correct envelope
W - Letter going into wrong envelope

No#1. C - W - W - W
............in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3

Thats the only explanation i was able to understand =(
I'm worried because if I don't learn the combinatrics formula way of doing it I might make a logical inconsistency while solving under a time limit.
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  15 Sep 2013, 11:23
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We are trying to find the probability of 1R3W.

Probability = number of ways to get 1R3W/number of ways total

number of ways total is 4! = 24. Imagine stuffing envelopes randomly. Stacy can put any of 4 letters into the first envelope, any of the remaining 3 into the next, either of the remaining 2 into the next, and has no choice to make on the last, or 4*3*2*1.

number of ways to get 1R3W : She could fill the first envelope with the right letter (1 way), then put either of the 2 wrong remaining letters in the next (2 ways), then put a wrong letter in the next (1 way). That's 1*2*1*1 = 2.

But since it doesn't have to be the first envelope that has the Right letter, it could be any of the 4 envelopes (i.e. we could have RWWW, WRWW, WWRW, WWWR), the total ways to get 1R3W is 4*2 = 8.

Probability is 8/24 = 1/3.
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Combinatorics and Probability [#permalink]  17 Jul 2010, 09:46
Hello there, good day

I'd love to hear your thoughts on this problem here below, I encountered it off one of my CAT practice test from the official prep

I'll write out the question as I remember it, so accurate meaning but not word for word, I also don't remember all the answer choices so I'll leave those out. Here goes

"You have 4 different letters to be sent to different addresses and 4 envelops with the different addresses printed on them. What is the probability that one letter is send to the wrong address"

I have confusion on several parts, but looking at it step by step, this is my logic

1- Probability is the number of outcomes which give us our criteria over the total number of outcomes. The total number of outcomes is 4! since this is essentially an ordered list, so the next question is, how many of those 4! outcomes satisfy the criteria of one letter put in the wrong envelope

2- If one letter is put in the wrong envelope, then at least another letter is in the wrong envelop as well. So this is equivalent to : Wrong/Wrong/*/*

This is where I start to get confused a bit, should we just simply count all the combinations in which there are at least two wrongs ? or should we pin/glue those two wrong and count only for the remaining 2! choices ? And if we do that, do we further need to multiply that result by 4 to account for circular list ?

I think the major points of confusion for me are :
a- I don't recall the question saying exactly one wrong, and I think that's impossible anyways, given that there are 4 distinct letters and 4 envelopes
b- I am quite confused about the "circular" logic list, if anything i'd really appreciate it if you could answer even just this
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Re: Combinations/Permutations [#permalink]  17 Jul 2010, 10:38
Expert's post
Merging your topic with the one with correct question.

It's not possible to assign only one letter to wrong envelope. Check this for all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letters%20envelopes

Hope it helps.
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Re: Combinations/Permutations [#permalink]  01 Feb 2012, 10:00
Hi, this is what I did:
Total possible outcomes are 4! = 24
Desired outcome(1 letter in correct envelope) = 4C1 (selecting 1 letter out of 4 to put in right envelope)*2 (then remaining 3 letters can be put in 3 envelopes in 2 wrong ways) = 8

Therefore probability of 1 letter in correct envelope is 8/24 = 1/3.
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  03 Apr 2012, 12:29
BN1989 wrote:
What if two letters are placed correctly, what's the probability of that? 1/6?

@pkonduri what's the general forumula for finding the derangements when none is placed correctly?

to answer your question if None is placed correctly, we need to calculate D(4) as 4 is the total number of letters that could possibley be placed in incorrect envelopes . the same formula(D(n) applies for all the cases
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  04 Aug 2012, 05:20
raghupara wrote:
Bunuel wrote:
BN1989 wrote:
What if two letters are placed correctly, what's the probability of that? 1/6?

@pkonduri what's the general forumula for finding the derangements when none is placed correctly?

Check this for all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html

Hope it helps.

Hi,

Is my understanding correct?

Total # of ways - 4! = 24.

Desired:
A-Mapped to the only correct address----------x 1 way only
B-Mapped to other two incorrect addresses - x 2 ways
C-Mapped to other two incorrect addresses - x 2 ways
D-Mapped to other two incorrect addresses - x 2 ways

Therefore, 1*2*2*2/24 = 1/3.?

Thanks.

You are presenting your solution for exactly one correctly placed letter?
I guess, you wanted to find in how many ways B,C and D can be placed so that none of them is in the right envelope.
This cannot give 2*2*2 = 8, because the total number of possible placements for the three letters is 3! = 6 < 8.

In addition, what about the cases when only B is placed correctly? Only C? Only D?
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  04 Aug 2012, 11:07
Yes, the total # of possible placements for the three letters is 3! = 6 in general, but in this case when we deal with the probability of these letters not being mapped to the right address, we have to take out the right one and so it becomes 2! = 2/letter.
Also, I did not have to compute the same again for other letters, meaning, P(A alone mapped to the right address, but the rest not) = P(B alone mapped to right address while others are not)........ = 2*2*2/4! = 1/3?
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  04 Aug 2012, 21:36
EvaJager wrote:
raghupara wrote:
Yes, the total # of possible placements for the three letters is 3! = 6 in general, but in this case when we deal with the probability of these letters not being mapped to the right address, we have to take out the right one and so it becomes 2! = 2/letter.
Also, I did not have to compute the same again for other letters, meaning, P(A alone mapped to the right address, but the rest not) = P(B alone mapped to right address while others are not)........ = 2*2*2/4! = 1/3?

(1) Using combinatorics:
To place one letter correctly (say A) - 1 possibility
The second letter to misplace it - 2 possibilities
The third letter to misplace it - 1 possibility
The fourth letter will be surely misplaced - just 1 possibility left
Consider the above scenario 4 times - 4 different possibilities for the correctly placed letter (A, B, C, or D can be placed correctly)
Therefore, the probability for exactly one letter placed correctly is 4 * 2 * 1 * 1/4! = 8/24 = 1/3

(2) Using probabilities:
To place one letter correctly (say A) - 1/4
The second letter to misplace it - 2/3
The third letter to misplace it - 1/2
The fourth letter will be surely misplaced - 1/1
So, the probability of a certain letter to be placed correctly and all the others to be misplaced is
(1/4)*(2/3)*(1/2) *(1/1) = 1/12
Again, consider the above scenario 4 times - it gives 4*(1/12) = 1/3

In your answer there are no probabilities (probabilities are expressed as fractions, between 0 and 1 inclusive).
Using the fundamental formula for probabilities (the number of desired outcomes / the total number of possible outcomes) again, your explanation isn't correct.
1*2*2*2 doesn't match any suitable scenario for placing exactly one later correctly and misplace all the other three.
Pay attention to getting the correct answer by using the correct reasoning.

Hey sure,
Thanks so many.. I got it now.
Kudos to you...

In fact I'm pathetically wrong...
The first time when I have mapped A to 1, I consumed that mapping and took only the remaining 3 in account and said only two were misfits therefore 2! = 4. Now, this act has to be for next letter(say B). So, for computing the placement of C, I should have exhausted/consumed that mapping of B as well. But i didn't and continued to use 2! for both C and D as well, which is way wrong !
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]  05 Aug 2012, 09:19
In a previous thread, IanStewart posted a solution to this question:

"One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3. "

I would like to point out that, although the final result is correct (1/3), the explanation is not correct.
"One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left." This assumption is nowhere reflected in the chain of computations.
In the above computations, 2/3 is what is called a conditional probability, meaning that under the assumption that a certain letter is placed correctly, then the probability of placing one of the remaining 3 letters incorrectly is 2/3, then the probability of placing one of the remaining two letters incorrectly after two were already placed, one correctly and one incorrectly, etc.

The correct chain should be: 1/4 for the probability of a certain letter to be placed correctly, 2/3 for one of the remaining three placed incorrectly, 1/2 for one of the remaining two to be misplaced, and 1/1 for the last one which will be surely misplaced. This gives (1/4)*(2/3)*(1/2)*(1/1)=1/12 for the probability of exactly one particular letter to be placed correctly.
Since there are four possibilities to chose the one letter placed correctly, the required probability is 4*(1/12) = 1/3.

There is a collection of questions with all the possible scenarios for placing the 4 letters:
letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letters%20envelopes

Can we use a similar reasoning used by IanStewart (for exactly 1 correctly placed letter) to find the probability of placing exactly 2 letters correctly?
The reasoning would go like this: consider two letters placed correctly, doesn't matter which ones, then the probability of placing one of the remaining letters incorrectly is 1/2, and then the fourth letter incorrectly is 1/1. Can we conclude that 1/2 is the probability of placing exactly two letters correctly?
The correct answer is 1/4. So, what's wrong with this reasoning?

I claim, we have to take into account the probability of placing a given pair of letters correctly. This probability is 1/12, since the probability of one of the letters to be placed correctly is 1/4, then the other one to be placed correctly is 1/3, which gives the (1/4)*(1/3) = 1/12 (or, either only the two letters are placed correctly or all four are, which means a probability of 2/24 = 1/12).
Then, the probability of one of the remaining two letters to be misplaced is 1/2, and the last one is for sure going to be misplaced (probability 1/1).
Since there are 4C2=(4*3)/2 = 6 possibilities to chose a pair of letters from the given 4 letters, the probability of placing exactly any two letters correctly is given by (1/12) * (1/2) * 6 = 1/4, which is the correct answer.

My point is: all the assumptions should be reflected in the computations, and first of all, we should make all the necessary and correct assumptions, regardless whether we are dealing directly with probabilities or counting number of possibilities. Easier said than done...but we have to try :O)
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Re: 4 letters 4 addresses probablity problem [#permalink]  17 Sep 2012, 22:39
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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] 17 Sep 2012, 22:39

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