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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

31 Aug 2013, 06:35

Expert's post

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... _________________

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

Show Tags

15 Sep 2013, 12:23

1

This post received 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.

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?

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?

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?

Please advice 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 - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\)

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?

Please advice 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 !!!

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) 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

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 ?

could you please help .

thanks lucky _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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:

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...