It is currently 22 Oct 2017, 20:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Tanya prepared 4 different letters to be sent to 4 different

Author Message
TAGS:

### Hide Tags

Manager
Affiliations: CFA L3 Candidate, Grad w/ Highest Honors
Joined: 03 Nov 2007
Posts: 131

Kudos [?]: 136 [6], given: 9

Location: USA
Schools: Chicago Booth R2 (WL), Wharton R2 w/ int, Kellogg R2 w/ int
WE 1: Global Operations (Futures & Portfolio Financing) - Hedge Fund (\$10bn+ Multi-Strat)
WE 2: Investment Analyst (Credit strategies) - Fund of Hedge Fund (\$10bn+ Multi-Strat)
Tanya prepared 4 different letters to be sent to 4 different [#permalink]

### Show Tags

11 Oct 2009, 11:51
6
KUDOS
52
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

43% (01:09) correct 57% (01:20) wrong based on 1150 sessions

### HideShow timer Statistics

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

Kudos [?]: 136 [6], given: 9

Math Expert
Joined: 02 Sep 2009
Posts: 41912

Kudos [?]: 129371 [15], given: 12197

Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

11 Oct 2009, 12:12
15
KUDOS
Expert's post
21
This post was
BOOKMARKED
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

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)
(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
_________________

Kudos [?]: 129371 [15], given: 12197

Manager
Joined: 13 Oct 2009
Posts: 55

Kudos [?]: 78 [30], given: 2

Location: New York, NY
Schools: Columbia, Johnson, Tuck, Stern
Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

10 Nov 2009, 10:04
30
KUDOS
10
This post was
BOOKMARKED
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

Kudos [?]: 78 [30], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 41912

Kudos [?]: 129371 [1], given: 12197

Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

10 Nov 2009, 10:50
1
KUDOS
Expert's post
You can also check the topic below, with almost all possible scenarios for this problem:

letter-arrangements-understanding-probability-and-combinats-84912.html?highlight=Tanya
_________________

Kudos [?]: 129371 [1], given: 12197

Intern
Joined: 18 Mar 2012
Posts: 47

Kudos [?]: 270 [0], given: 117

GPA: 3.7
Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

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

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)
(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!

Kudos [?]: 270 [0], given: 117

Manager
Status: Trying.... & desperate for success.
Joined: 17 May 2012
Posts: 73

Kudos [?]: 106 [1], given: 61

Location: India
Schools: NUS '15
GPA: 2.92
WE: Analyst (Computer Software)
Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

18 Mar 2013, 07:55
1
KUDOS
1
This post was
BOOKMARKED
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

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)
(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

Kudos [?]: 106 [1], given: 61

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17396 [3], given: 232

Location: Pune, India
Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

18 Mar 2013, 21:33
3
KUDOS
Expert's post
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

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)
(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/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17396 [3], given: 232

Intern
Joined: 07 Aug 2012
Posts: 20

Kudos [?]: 1 [0], given: 18

Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

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 ?

Kudos [?]: 1 [0], given: 18

Math Expert
Joined: 02 Sep 2009
Posts: 41912

Kudos [?]: 129371 [0], given: 12197

Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

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.
_________________

Kudos [?]: 129371 [0], given: 12197

Intern
Joined: 18 Jan 2014
Posts: 12

Kudos [?]: 18 [2], given: 14

GMAT 1: 640 Q49 V28
GPA: 3.5
WE: Operations (Energy and Utilities)
Re: 4 letters & 4 envelopes [#permalink]

### Show Tags

16 Jun 2014, 10:19
2
KUDOS
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)
(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.

Kudos [?]: 18 [2], given: 14

Intern
Joined: 31 Aug 2013
Posts: 10

Kudos [?]: 6 [1], given: 1

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

### Show Tags

10 Aug 2014, 23:25
1
KUDOS
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!]

Kudos [?]: 6 [1], given: 1

Intern
Joined: 09 Aug 2016
Posts: 3

Kudos [?]: [0], given: 3

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

### Show Tags

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.

Kudos [?]: [0], given: 3

Senior Manager
Joined: 25 Feb 2013
Posts: 436

Kudos [?]: 198 [0], given: 31

Location: India
GPA: 3.82
Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

### Show Tags

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

Kudos [?]: 198 [0], given: 31

Intern
Joined: 06 Feb 2017
Posts: 2

Kudos [?]: 3 [0], given: 1

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

### Show Tags

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

Kudos [?]: 3 [0], given: 1

BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2212

Kudos [?]: 845 [0], given: 595

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

### Show Tags

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.

_________________

Give me a hell yeah ...!!!!!

Kudos [?]: 845 [0], given: 595

Director
Joined: 17 Dec 2012
Posts: 608

Kudos [?]: 518 [0], given: 16

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

### Show Tags

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.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/regularcourse.php

Pay After Use
Standardized Approaches

Kudos [?]: 518 [0], given: 16

Intern
Joined: 05 Jun 2016
Posts: 32

Kudos [?]: 1 [0], given: 742

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

### Show Tags

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.

Kudos [?]: 1 [0], given: 742

BSchool Forum Moderator
Joined: 17 Jun 2016
Posts: 400

Kudos [?]: 188 [0], given: 198

Location: India
GMAT 1: 720 Q49 V39
GPA: 3.65
WE: Engineering (Energy and Utilities)
Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

### Show Tags

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

Refer to solution in the picture
Attachments

WhatsApp Image 2017-06-25 at 10.14.43 PM.jpeg [ 27.78 KiB | Viewed 8825 times ]

_________________

Kudos [?]: 188 [0], given: 198

Re: Tanya prepared 4 different letters to be sent to 4 different   [#permalink] 25 Jun 2017, 11:16
Display posts from previous: Sort by