Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Aug 2016, 10:00

### 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 4 different addresses

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

BSchool Forum Moderator
Joined: 27 Aug 2012
Posts: 1192
Followers: 123

Kudos [?]: 1296 [1] , given: 142

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

### Show Tags

31 Aug 2013, 00:33
1
KUDOS
1
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

45% (02:08) correct 55% (01:28) wrong based on 626 sessions

### HideShow timer Statistics

Hi Guys,
Can anybody shot an explanation to my query raised above ?

I'd much appreciate if any Quant expert kindly explain this issue to me!
[Reveal] Spoiler: OA

_________________
Intern
Joined: 09 Jul 2012
Posts: 11
Followers: 0

Kudos [?]: 3 [1] , given: 6

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

### Show Tags

15 Sep 2013, 12:23
1
KUDOS
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.
Math Expert
Joined: 02 Sep 2009
Posts: 34438
Followers: 6257

Kudos [?]: 79463 [0], given: 10018

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

### Show Tags

31 Aug 2013, 06:35
Expert's post
1
This post was
BOOKMARKED
bagdbmba wrote:
Hi Guys,
Can anybody shot an explanation to my query raised above ?

I'd much appreciate if any Quant expert kindly explain this issue to me!

This case is solved in the following way:
Counting all incorrect, or 0 correct:
P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.

Don;t know what I can add to that...
_________________
BSchool Forum Moderator
Joined: 27 Aug 2012
Posts: 1192
Followers: 123

Kudos [?]: 1296 [0], given: 142

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

### Show Tags

31 Aug 2013, 10:51
Got it Sir!
It's clear now...

this

_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 54

Kudos [?]: 513 [0], given: 355

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

### Show Tags

21 May 2014, 05:35
Aximili85 wrote:
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.

Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Cheers!
J
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 54

Kudos [?]: 513 [0], given: 355

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

### Show Tags

21 May 2014, 05:35
Aximili85 wrote:
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.

Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Cheers!
J
Manager
Joined: 13 Feb 2011
Posts: 104
Followers: 0

Kudos [?]: 23 [0], given: 3358

Tanya prepared 4 different letters to 4 different addresses [#permalink]

### Show Tags

20 Sep 2014, 17:21
jlgdr wrote:
Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Cheers!
J

I actually had the same question and was about to post a reply on this thread seeking same but while typing it had the aha moment.

Yes, even when one calculates the probability of each scenario separately, the total will be $$\frac{1}{3}$$.
#1 - C W W W - $$\frac{1}{4}*\frac{2}{3}*\frac{1}{2}*1=\frac{1}{12}$$
#2 - W C W W - $$\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}$$ (Note for the first W, desired outcomes are only two, the 3rd and 4th envelopes, and doesn't include its own envelope and the envelope for the second letter as that needs to have the correct one. Same explanation plays in scenarios #3 and #4)
#3 - W W C W - $$\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}$$
#4 - W W W C - $$\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}$$
Chat Moderator
Joined: 19 Apr 2013
Posts: 748
Concentration: Strategy, Healthcare
Schools: Sloan '18 (A)
GMAT 1: 730 Q48 V41
GPA: 4
Followers: 10

Kudos [?]: 161 [0], given: 537

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

### Show Tags

01 Dec 2014, 03:09
Dienekes wrote:
jlgdr wrote:
Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Cheers!
J

I actually had the same question and was about to post a reply on this thread seeking same but while typing it had the aha moment.

Yes, even when one calculates the probability of each scenario separately, the total will be $$\frac{1}{3}$$.
#1 - C W W W - $$\frac{1}{4}*\frac{2}{3}*\frac{1}{2}*1=\frac{1}{12}$$
#2 - W C W W - $$\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}$$ (Note for the first W, desired outcomes are only two, the 3rd and 4th envelopes, and doesn't include its own envelope and the envelope for the second letter as that needs to have the correct one. Same explanation plays in scenarios #3 and #4)
#3 - W W C W - $$\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}$$
#4 - W W W C - $$\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}$$

Bunuel, can you check this post, pls. I think there are mistakes in calculations of 2nd, 3rd, and 4th conditions. For the first W I think there must 3/4. Am I right?
_________________

If my post was helpful, press Kudos. If not, then just press Kudos !!!

Director
Joined: 07 Aug 2011
Posts: 582
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27
Followers: 3

Kudos [?]: 338 [0], given: 75

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

### Show Tags

14 Mar 2015, 10:53
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

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.

Hi Bunuel,
what if there were 5 letters and we were asked to find the same probability as in the original question , how would we calculate these possible wrong arrangements using combitronics ?
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
ABCDE
Total arrangements for BCDE=4!
case 1: only 2 wrong : 4
case 2: three wrong : X
so all wrong will be 4!-4 -X ?

thanks
lucky
_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the to appreciate my post !!

VP
Joined: 09 Jun 2010
Posts: 1331
Followers: 3

Kudos [?]: 96 [0], given: 768

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

### Show Tags

07 Jun 2015, 02:15
hard question, but if you do this two time, second time one month after first time, you can remember

I try to do this one for 3 times.
now I master this problem
Intern
Joined: 10 Jun 2015
Posts: 23
Concentration: Marketing, Technology
GMAT 1: 490 Q47 V13
Followers: 0

Kudos [?]: 0 [0], given: 9

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

### Show Tags

21 Jan 2016, 12:18
GMATGuruNY 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

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.

The correct answer is
[Reveal] Spoiler:
D
.

I find your solution, the best among all. Thanks.
_________________

Abhilash
Learn, Apply, Fail, Analyse, and Practice.

Consider +1 Kudos if you find my post helpful.

Re: Tanya prepared 4 different letters to 4 different addresses   [#permalink] 21 Jan 2016, 12:18
Similar topics Replies Last post
Similar
Topics:
94 Tanya prepared 4 letters to be sent to 4 different addresses. For each 23 31 Dec 2009, 21:22
1 Tanya prepared 4 different letters to be sent to 4 different 1 09 Jul 2012, 08:00
61 Tanya prepared 4 different letters to be sent to 4 different 12 11 Oct 2009, 11:51
3 Tanya prepared 4 different letters to 4 different addresses. 9 08 Sep 2009, 09:28
20 Tanya prepared 4 letters to be sent to 4 different 11 15 Oct 2006, 14:12
Display posts from previous: Sort by

# Tanya prepared 4 different letters to 4 different addresses

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.