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

 It is currently 09 Feb 2016, 02:43

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 letters to be sent to 4 different

Author Message
TAGS:
Intern
Joined: 08 Aug 2006
Posts: 2
Followers: 0

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

Tanya prepared 4 letters to be sent to 4 different [#permalink]  15 Oct 2006, 13:12
1
KUDOS
00:00

Difficulty:

95% (hard)

Question Stats:

39% (02:54) correct 61% (01:37) wrong based on 126 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: tanya-prepared-4-different-letters-to-4-different-addresses-88626.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Jul 2014, 11:20, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
 Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
Manager
Joined: 07 Jun 2006
Posts: 111
Followers: 2

Kudos [?]: 12 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  15 Oct 2006, 19:07
Is it C - 1/4??

Desired Results - 3! * 4C1
Total possible ways - 4! * 4C1

p(A) = 3! * 4C1/4! * 4C1 = 6/24 = 1/4

Correct me if I am wrong.
VP
Joined: 25 Jun 2006
Posts: 1173
Followers: 2

Kudos [?]: 88 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  15 Oct 2006, 19:33
D. i got 1/3.

total #: 4!

desired #:
Anyone can be correct. so prepare to time the following by 4:
the rest of the 3 can not be correct, i.e. 3 must be all wrong.
out of 3 permutations, u have 2 choices that make all 3 letters wrong.

so:

4*2/4! = 1/3
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 33 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  23 Oct 2006, 00:52
hi tennis_ball, didn't get your explanation. why is it 1/3? is there another way you can explain? sorry, this my weak point. [/u]
_________________

Impossible is nothing

Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  23 Oct 2006, 01:37
1
This post was
BOOKMARKED
I got D as well

if only one is correct:

the probability = correct*incorrect*incorrect*incorrect

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

since there are 4 letters, there are 4 possibilities:
1/12*4C1=1/3
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 33 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  23 Oct 2006, 02:32
sorrry guys... still not getting thru my brain... any other (possibly more basic/ step by step) explanation?
_________________

Impossible is nothing

VP
Joined: 15 Jul 2004
Posts: 1473
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
Followers: 19

Kudos [?]: 133 [14] , given: 13

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  23 Oct 2006, 08:31
14
KUDOS
1
This post was
BOOKMARKED
Hermione wrote:
sorrry guys... still not getting thru my brain... any other (possibly more basic/ step by step) explanation?

OK in that case let me see if I can offer some help -

I hope you are familiar with basic probability fundas -

Let's say you have just ONE letter and TWO envelopes ONE of which is correctly addressed and the other addressed incorrectly.

What's the probability of putting the Letter in the correctly addressed envelope -

To answer this question - we see IN HOW MANY WAYS can the letter be put into the envelope - you could put it (assuming you don't know which envelope is which) in either of the two - so in total you have TWO ways of shoving the letter in. However, there's only ONE way in which it can go into the correctly addressed envelope -

so 1/2 is the prob of putting in correct envelope. This is easy.

Now in our current problem - let's say we have just ONE letter but FOUR envelopes. Only one of these envelopes has the address corresponding to the letter. The remaining three envelopes are incorrectly addressed.

So the probability that you will put the letter correctly is 1/4. Right?

What happens if i ask you the reverse question? what is the prob for putting it in the incorrect envelope. Suddenly you have three envs that are incorrect so you can put the letter incorrectly with a prob of 3/4. Right?

The whole problem can be broken down into Four Events that will fulfill the requirement of the question

Event 1 - E1

We know that prob of putting ONE Letter correctly is 1/4. Now once ONE letter has been put CORRECTLY, what are you LEFT with? You are left with THREE ENVELOPES and the remaining THREE letters. Since the one letter has been put correctly (though technically we have just calculated the PROBABILITY that the first letter goes into the correct envelope) we have the remaining THREE Letters and THREE envelopes.

Event 2 - E2

Let's take letter number 2 now - what is the probability that it LANDS in the INCORRECT envelope. Again by the same logic as above - there are 3 envelopes remaining out of which ONLY ONE has the correct address for LETTER number 2. The remaining 2 have INCORRECT address and LETTER NUMBER 2 could go in either of these 2 to meet our condition.
Thus the probability of this event is 2/3

So till now what we have done is -

we have calculated the prob of shoving Letter number 1 in correct env -- 1/4

we have calculated the prob of shoving Letter number 2 in INcorrect env --- 2/3

Event 3 - E3

Now let's take letter number 3 - again according to question we want to shove this in the WRONG envelope. There are 2 remaining envelopes and hence the prob of shoving this in the wrong env (or equally in the RIght env) is 1/2.

Finally we come to event E4 - the Letter number 4. This has only one way of going in so its probability of being put into the WRONG envelope is 1.

ok so we can see that our grand event is actually a combination of FOUR EVENTS happening - each with a probability of its own. So to calculate the total probability of the Grand Event itself we just multiply the individual probabilities since each event happens INDEPENDENTLY of each other

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

However at this point - I must introduce one last element in this question -since there are FOUR Letters - what we saw above was JUST ONE SEQUENCE of events leading to the desired result.

If we arbitrarily call the letters L1 thru L4, and let's say the above was an example in which we started by Picking up Letter L1 and worked thru the remaining letters, we could have equally well started out with letter L2 or L3 or L4.

Thus since each of these events ARE MUTUALLY EXCLUSIVE, meaning THEY CAN NEVER HAPPEN ALL THE SAME TIME BUT ONLY ONE LETTER AT A TIME, to calculate the TOTAL PROBABILITY of we will add the individual probabilities 1/12 + 1/12 + 1/12 + 1/12 which works out to 1/3.

Phew!!
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 33 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  24 Oct 2006, 08:05
thanks for the very thorough explanation, dwivedys... i owe you appreciate your help
_________________

Impossible is nothing

Manager
Joined: 04 May 2006
Posts: 173
Location: paris
Followers: 1

Kudos [?]: 14 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  24 Oct 2006, 11:55
Trying to solve this question for the first time, solution above is still not clear consider the following situation:

What about if the first letter is not shoved right?
Don't we need to consider these scenarii?

R= shoving letter x on the right envelope
NR= not shoving letter x on the right envelope

NR*R*R*R=1/4*2/3*1/2*1/1=1/12
R*NR*R*R=3/4*1/3*1/2*1=1/8
R*R*R*NR=3/4*2/3*1/2*1=1/4
_________________

time is not on my side

Director
Joined: 17 Sep 2005
Posts: 924
Followers: 3

Kudos [?]: 53 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  04 Oct 2007, 20:16
dwivedys wrote:
Hermione wrote:
sorrry guys... still not getting thru my brain... any other (possibly more basic/ step by step) explanation?

OK in that case let me see if I can offer some help -

I hope you are familiar with basic probability fundas -

Let's say you have just ONE letter and TWO envelopes ONE of which is correctly addressed and the other addressed incorrectly.

What's the probability of putting the Letter in the correctly addressed envelope -

To answer this question - we see IN HOW MANY WAYS can the letter be put into the envelope - you could put it (assuming you don't know which envelope is which) in either of the two - so in total you have TWO ways of shoving the letter in. However, there's only ONE way in which it can go into the correctly addressed envelope -

so 1/2 is the prob of putting in correct envelope. This is easy.

Now in our current problem - let's say we have just ONE letter but FOUR envelopes. Only one of these envelopes has the address corresponding to the letter. The remaining three envelopes are incorrectly addressed.

So the probability that you will put the letter correctly is 1/4. Right?

What happens if i ask you the reverse question? what is the prob for putting it in the incorrect envelope. Suddenly you have three envs that are incorrect so you can put the letter incorrectly with a prob of 3/4. Right?

The whole problem can be broken down into Four Events that will fulfill the requirement of the question

Event 1 - E1

We know that prob of putting ONE Letter correctly is 1/4. Now once ONE letter has been put CORRECTLY, what are you LEFT with? You are left with THREE ENVELOPES and the remaining THREE letters. Since the one letter has been put correctly (though technically we have just calculated the PROBABILITY that the first letter goes into the correct envelope) we have the remaining THREE Letters and THREE envelopes.

Event 2 - E2

Let's take letter number 2 now - what is the probability that it LANDS in the INCORRECT envelope. Again by the same logic as above - there are 3 envelopes remaining out of which ONLY ONE has the correct address for LETTER number 2. The remaining 2 have INCORRECT address and LETTER NUMBER 2 could go in either of these 2 to meet our condition.
Thus the probability of this event is 2/3

So till now what we have done is -

we have calculated the prob of shoving Letter number 1 in correct env -- 1/4

we have calculated the prob of shoving Letter number 2 in INcorrect env --- 2/3

Event 3 - E3

Now let's take letter number 3 - again according to question we want to shove this in the WRONG envelope. There are 2 remaining envelopes and hence the prob of shoving this in the wrong env (or equally in the RIght env) is 1/2.

Finally we come to event E4 - the Letter number 4. This has only one way of going in so its probability of being put into the WRONG envelope is 1.

ok so we can see that our grand event is actually a combination of FOUR EVENTS happening - each with a probability of its own. So to calculate the total probability of the Grand Event itself we just multiply the individual probabilities since each event happens INDEPENDENTLY of each other

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

However at this point - I must introduce one last element in this question -since there are FOUR Letters - what we saw above was JUST ONE SEQUENCE of events leading to the desired result.

If we arbitrarily call the letters L1 thru L4, and let's say the above was an example in which we started by Picking up Letter L1 and worked thru the remaining letters, we could have equally well started out with letter L2 or L3 or L4.

Thus since each of these events ARE MUTUALLY EXCLUSIVE, meaning THEY CAN NEVER HAPPEN ALL THE SAME TIME BUT ONLY ONE LETTER AT A TIME, to calculate the TOTAL PROBABILITY of we will add the individual probabilities 1/12 + 1/12 + 1/12 + 1/12 which works out to 1/3.

Phew!!

This is Tutoring!!

- Brajesh
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 8184
Followers: 416

Kudos [?]: 111 [0], given: 0

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  07 Jul 2014, 10:44
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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 31287
Followers: 5349

Kudos [?]: 62227 [2] , given: 9444

Re: Tanya prepared 4 letters to be sent to 4 different [#permalink]  07 Jul 2014, 11:22
2
KUDOS
Expert's post
tealeaflin wrote:
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

$$P(C=1)=\frac{8}{24}=\frac{1}{3}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: tanya-prepared-4-different-letters-to-4-different-addresses-88626.html
_________________
Re: Tanya prepared 4 letters to be sent to 4 different   [#permalink] 07 Jul 2014, 11:22
Similar topics Replies Last post
Similar
Topics:
1 Tanya prepared 4 different letters to be sent to 4 different 1 09 Jul 2012, 07:00
8 4 different letters to be sent to 4 addresses. For each lett 6 31 Oct 2011, 02:41
84 Tanya prepared 4 different letters to 4 different addresses 30 31 Dec 2009, 20:22
46 Tanya prepared 4 different letters to be sent to 4 different 11 11 Oct 2009, 10:51
2 Tanya prepared 4 different letters to 4 different addresses. 9 08 Sep 2009, 08:28
Display posts from previous: Sort by