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:

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

Show Tags

11 Oct 2009, 11:51

3

This post received KUDOS

16

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

41% (02:12) correct
59% (01:29) wrong based on 582 sessions

HideShow timer Statictics

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?

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?

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

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:

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?

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

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!

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?

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

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,

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?

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

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!

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 ?

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. _________________

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

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]

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

Show Tags

10 Aug 2014, 23:25

1

This post received 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!]

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

Show Tags

11 Aug 2015, 09:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...