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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]

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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 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?

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

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Re: 4 letters & 4 envelopes [#permalink]

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robertrdzak wrote:
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 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?

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

[Reveal] Spoiler: OA
D) 1/3


Could someone please explain a good way to go about solving this please?


Total # of ways of choosing envelopes=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)
ADBC(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)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html
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Re: 4 letters & 4 envelopes [#permalink]

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You can solve this the traditional way:

Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1

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

Multiply by 4, representing the four letters in the correct envelope:

1/12 * 4 = 4/12 = 1/3

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Re: 4 letters & 4 envelopes [#permalink]

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Re: 4 letters & 4 envelopes [#permalink]

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New post 17 Mar 2013, 10:19
Bunuel wrote:
robertrdzak wrote:
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 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?

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

[Reveal] Spoiler: OA
D) 1/3


Could someone please explain a good way to go about solving this please?


Total # of ways of choosing envelopes=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)
ADBC(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)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html



Hi,

I do not get why we are multiplying by 4? Is it because we have to repeat the process described above for each of A, B, C and D as we do not know which one will be the right one?
Thanks!

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Re: 4 letters & 4 envelopes [#permalink]

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New post 18 Mar 2013, 07:55
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alex1233 wrote:
Bunuel wrote:
robertrdzak wrote:
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 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?

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

[Reveal] Spoiler: OA
D) 1/3


Could someone please explain a good way to go about solving this please?


Total # of ways of choosing envelopes=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)
ADBC(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)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html



Hi,

I do not get why we are multiplying by 4? Is it because we have to repeat the process described above for each of A, B, C and D as we do not know which one will be the right one?
Thanks!


Hi,
probability of ONE letter being in correct envelope and rest of the other 3 being in in-correct envelope is [1/4] * [2/3 * 1/2 * 1] = 1/12

Say there are 4 letters ABCD, then per above scenario, we are just finding the probability of just one letter A. We have B,C & D as well.
So the probability of letters B,C&D to individually having a chance to put in correct envelope is,

4 * 1/12 = 1/3

Hence the answer

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Re: 4 letters & 4 envelopes [#permalink]

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New post 18 Mar 2013, 21:33
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alex1233 wrote:
Bunuel wrote:
robertrdzak wrote:
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 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?

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

[Reveal] Spoiler: OA
D) 1/3


Could someone please explain a good way to go about solving this please?


Total # of ways of choosing envelopes=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)
ADBC(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)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html



Hi,

I do not get why we are multiplying by 4? Is it because we have to repeat the process described above for each of A, B, C and D as we do not know which one will be the right one?
Thanks!


Check out all three letter and four letter scenarios here:
http://www.veritasprep.com/blog/2011/12 ... envelopes/
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Re: 4 letters & 4 envelopes [#permalink]

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New post 30 Nov 2013, 03:16
Wayxi wrote:
You can solve this the traditional way:

Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1

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

Multiply by 4, representing the four letters in the correct envelope:

1/12 * 4 = 4/12 = 1/3



Dear Wayxi ...why u take probability for second letter as 2/3 ..i do understand 3 but confused about 2 in numerator . Isnt it should be 1/3 ?

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Re: 4 letters & 4 envelopes [#permalink]

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New post 30 Nov 2013, 04:19
archit wrote:
Wayxi wrote:
You can solve this the traditional way:

Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1

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

Multiply by 4, representing the four letters in the correct envelope:

1/12 * 4 = 4/12 = 1/3



Dear Wayxi ...why u take probability for second letter as 2/3 ..i do understand 3 but confused about 2 in numerator . Isnt it should be 1/3 ?


When one letter is in right envelope, there are 3 left. The probability that the second letter gets in WRONG is 2/3.
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Re: 4 letters & 4 envelopes [#permalink]

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Total # of ways of choosing envelopes=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)
ADBC(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)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html[/quote]


Hi,

I do not get why we are multiplying by 4? Is it because we have to repeat the process described above for each of A, B, C and D as we do not know which one will be the right one?
Thanks![/quote]

Check out all three letter and four letter scenarios here:
http://www.veritasprep.com/blog/2011/12 ... envelopes/[/quote]



THANKS A LOT ! THE EXPLANATION IS NOT ONLY CONVINCING BUT ALSO EASY TO GRASP.

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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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robertrdzak wrote:
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 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?

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



We can also do this question by derangement method:

1. First choose one of the letters and put it in right envelope:
That can be done in -> 4C1= 4 ways.

2. Now we would derange the rest of 3 envelopes in :
3! (1/2! - 1/3!) = 2 ways
Finally the number of ways will be = statement 1 x statement 2= 4x2= 8 ways --------------- 3

We have sample space= 4! (number of ways of arranging 4 different letters) = 24 ways ---------------- 4

So the probability will be = statement 3/ statement 4 = 8/24= 1/3 (answer)

P.S. In general the number of ways of derangement of n things D(n)= n! [1/2! -1/3!+1/4!- .....+ (-1)^n/n!]

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Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 06 Nov 2016, 15:46
Wayxi wrote:
You can solve this the traditional way:

Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1

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

Multiply by 4, representing the four letters in the correct envelope:

1/12 * 4 = 4/12 = 1/3


I think this method may have flaws. What if the second letter is in the right envelope? In this case, those probabilities would be:
Probability that first letter in the wrong envelope= 3/4
Probability that second letter in right envelope = 1/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1
3/4*1/3*1/2*1=1/8

I am confused. Please explain. Thank you.

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Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 11 Feb 2017, 11:06
allenmaxin wrote:
Wayxi wrote:
You can solve this the traditional way:

Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1

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

Multiply by 4, representing the four letters in the correct envelope:

1/12 * 4 = 4/12 = 1/3


I think this method may have flaws. What if the second letter is in the right envelope? In this case, those probabilities would be:
Probability that first letter in the wrong envelope= 3/4
Probability that second letter in right envelope = 1/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1
3/4*1/3*1/2*1=1/8

I am confused. Please explain. Thank you.


Hi,
In your approach, you are taking into consideration arrangement of 4 letters into 1st, 2nd, 3rd & 4th envelop, whereas for this question order does not matter.
So, if you have 4 letters, A B C D then,
probability of A getting into correct envelop will be = 1/4*2/3*1/2*1 = 1/12
Probability of B,C,D getting into correct envelop will be same
Hence probability of only 1 letter getting into correct envelop will be = 1/12+1/12+1/12+1/12 = 4*1/12 = 1/3

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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 24 Apr 2017, 18:41
Wayxi wrote:
You can solve this the traditional way:

Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1

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

Multiply by 4, representing the four letters in the correct envelope:

1/12 * 4 = 4/12 = 1/3


For anyone wondering about this solution, it is actually correct only by coincidence. This method will fail when the number of letters increased to 5, and is therefore an incorrect formula.

"There are issues with this calculation. it happens to hit upon the correct value at the end, but that's a total coincidence.

i agree with the first two probabilities: the probability that letter a goes into envelope a is indeed 1/4, and the probability if that happens that the letter b goes into an envelope other than b is 2/3.
however, it's downhill from there: if letter b actually went into envelope c, then the probability of letter c not going into envelope c is 1. the probability is only 1/2 (as you've stated) if letter b winds up in envelope d.
similarly, the final probability is either 0 or 1, depending on whether the last envelope remaining is envelope d or not. if letter b goes in envelope c and letter c goes in envelope b (fulfilling all of your conditions), then letter d is stuck going into envelope d, making that last probability 0.

so, if you're going to go this route, you're stuck with doing the following:
* first 2 steps = same as you have them now
* 3rd step = 2 branches of a probability tree, depending on whether envelope c is still available (vs. whether it was used for letter b)
* 4th step = 2 branches off EACH of those prior 2 branches, depending on whether envelope d is still available (vs. whether it was used for letter b or c)"

If you would like to read further, see Manhattan Prep's forum post on this question.

This is a quote from Ron Purewal

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Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 28 Apr 2017, 01:24
Here is what i did on this question =>

P(E) =>
(1/4) * (2/4) * (1/2) *(1/1) + (2/4)*(1/3)*(1/2)*(1/1) + (2/4)*(1/3)*(1/2)*(1/1)+(2/4)*(1/3)*(1/2)*(1/1)

1/12 + 1/12 + 1/12 + 1/12

4/12 => 1/3

SMASH THAT D.

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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 31 May 2017, 03:55
@
robertrdzak wrote:
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 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?

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

1. Total number of arrangements is 24
2. List out the arrangements partly, if letter 1 is in the first envelope
They are 1234, 1243, 1324,1342, 1423, 1432
3. Out of the six above cases, 2 cases satisfy the condition. It will be the same for the other letters in the first envelope
4. So the probability is 8/24 which is 1/3.
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Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 15 Jun 2017, 05:11
This is how I did it. Please tell me if my method is right.

Let the 4 letters be ABCD. Let the 4 addresses/envelopes be EFGH.

Suppose the right letter-envelope combo is as follows: 1) A-E; (2) B-F; (3) C-G; (4)D-H . Thus, there are 4 total ways of putting the right letter into the right envelope.

Now, number of ways of getting the wrong letter envelope combo: (1)AF (2)AG (3) AH (4) BE (5)BG (6)BH , etc. You will have 3 +3 for each of letters C and D. Therefore, total ways of getting it wrong 12.

Now question asks the probability of getting 1 letter into the right envelope (i.e. any one of those 4 correct combo's) = 4/12 = 1/3.

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Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 25 Jun 2017, 11:16
robertrdzak wrote:
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 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?

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


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