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

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

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Updated on: 05 Sep 2019, 05:41
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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

Originally posted by tealeaflin on 15 Oct 2006, 14:12.
Last edited by Bunuel on 05 Sep 2019, 05:41, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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11 Oct 2009, 12:12
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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: http://gmatclub.com/forum/letter-arrang ... 84912.html

Hope it's clear.
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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10 Nov 2009, 10:04
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37
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
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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23 Oct 2006, 03:32
sorrry guys... still not getting thru my brain... any other (possibly more basic/ step by step) explanation?
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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23 Oct 2006, 09:31
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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!!
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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08 Sep 2009, 11:00
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For the Tanya question... Ans is 1/3

Lets take 4 letters L1 L2 L3 L4 and 4 envelopes E1 E2 E3 E4 ... L1 should go into E1 and so on...

The Question asks us to find the prob of only one letter going into the correct envelope, which means the other 3 go into wrong envelopes.

Initially, Lets find the Total no of ways of arranging 4 letters in 4 diff envelopes which is 4! = 24

so, L1 to go into E1(correct envelope)...Thus..... 1 choice
L2 can go into E3 or E4(wrong envelopes)........ 2 choices
L3 can go into only E2 or E4(wrong envelopes)...2 choices
L4 can go only to E2 or E3(wrong envelopes).....2 choices

Thus probability that only 1 letter will be put into the envelope with its correct address= 1*2*2*2/4!=8/24 => 1/3
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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08 Sep 2009, 11:17
ank wrote:
For the Tanya question... Ans is 1/3

Lets take 4 letters L1 L2 L3 L4 and 4 envelopes E1 E2 E3 E4 ... L1 should go into E1 and so on...

The Question asks us to find the prob of only one letter going into the correct envelope, which means the other 3 go into wrong envelopes.

Initially, Lets find the Total no of ways of arranging 4 letters in 4 diff envelopes which is 4! = 24

so, L1 to go into E1(correct envelope)...Thus..... 1 choice
L2 can go into E3 or E4(wrong envelopes)........ 2 choices
L3 can go into only E2 or E4(wrong envelopes)...2 choices
L4 can go only to E2 or E3(wrong envelopes).....2 choices

Thus probability that only 1 letter will be put into the envelope with its correct address= 1*2*2*2/4!=8/24 => 1/3

OA is 1/3....

I took little diff approach....please explain wht's wrong with it???

Probab of 1 letter to correct add ENV = 4C1 / 4! = 1/6, while answer is just double of this WHY???
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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12 Sep 2009, 13:24
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GODSPEED wrote:
OA is 1/3....

I took little diff approach....please explain wht's wrong with it???

Probab of 1 letter to correct add ENV = 4C1 / 4! = 1/6, while answer is just double of this WHY???

The number of ways to put exactly one letter in the right envelope is not equal to 4C1. There are 4C1 ways to choose which letter goes in the right envelope, but you then need to work out how many ways the remaining letters can be placed in the wrong envelope. For the first of these three letters, there are 2 wrong envelopes you could choose. Now, you still have one letter left which has its matching envelope unused; you must put this letter in the wrong envelope, so you have only 1 choice for this letter, and finally for the last letter you have only 1 choice for where to put it. So you have 2*1*1 = 2 ways to assign the remaining letters incorrectly, which is why you need to multiply your answer by 2.

I posted a slightly different solution to BTG a while ago, which I'll paste here:

One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3.
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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10 Nov 2009, 10:50
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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
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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19 Oct 2010, 11:45
10
5
1/3

Only 1 letter finds the correct envelope --- that can happen in 4 cases when each of the 4 letters fills into the correct envelope

So, 1st envelope: can be filled in 4 ways (i.e., 4 possible correct letters)

The other 3 go astray:

2nd envelope: can be filled in 2 ways (2 wrong letters)
3rd envelope: can be filled in 1 way (1 wrong letter)
4th and final envelope: can be filled in 1 way (1 wrong letter)

So number of desired events = 4*2*1*1 = 8

Total ways to place 4 letters in 4 envelopes = 4*3*2*1 = 24

Prob = Desired / Total = 8/24 = 1/3
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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19 Oct 2010, 12:07
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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.

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

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31 Oct 2011, 04:00
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Hi,

Can someone pls let me know what is the correct answer for below given problem?

4 different letters to be sent to 4 addresses. For each letter we have only one envelop with correct address. 4 letters are to be
put in 4 envelops randomly. What is the probability that only one letter will be put into envelop with correct address?
A) 1/24 B) 1/8 C) 1/4 D) 1/3 E) 3/8

Think of it this way:

There are 4 letters, R, G, B and Y and 4 envelopes, R, G, B and Y
Letter R should go in envelope R, letter G in envelope G etc
But if we are putting letters in randomly, we can arrange 4 letters in 4 envelopes in 4! = 24 ways (using basic counting principle. Check http://www.veritasprep.com/blog/2011/10 ... inatorics/)

Now let us select 1 correct combination out of the 4 in 4C1 = 4 ways
e.g. letter B is put in envelope B.

Now we have 3 letters and 3 corresponding envelopes. They must all go in incorrectly. In how many ways can this be done?
We can put in a letter incorrectly in envelope R in 2 ways (we can put in letter G or Y) e.g. we put letter G in envelope R. Now we have letters R and Y remaining.

Now, there is only 1 way of putting letters in G and Y. Envelope Y should not have letter Y so it must have letter R and envelope G must have the remaining letter Y.

Total ways of arranging such that only one letter goes in correctly = 4*2*1 = 8 ways

Probability of putting in only 1 letter correctly = 8/24 = 1/3
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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31 Oct 2011, 04:20
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Here is my take:

We have 4 different ways to get exactly one correct match.

Therefore 4xP(exactly one match)

4x(probability of getting 1st right)X(probability of getting 2nd wrong)X(probability of getting 3rd wrong)X(probability of getting 4th wrong)

Probability of getting 1st right= 1/4 Imagine we have picked a letter and need to pick its corresponding envelope. There is only one correct envelope out of 4 possible options.

Probability of getting 2nd wrong= 2/3 Now we pick the 2nd letter and are looking to get the wrong envelope. Out of the 3 total envelopes present 2 are wrong and one is its match.

Probability of getting 3rd wrong= 1/2 same logic as previous step. 2 envelopes remains out of which one is wrong the wrong one and we need to pick that.

Probability of getting 4th wrong=1 we don’t have any option now as we have already picked the remaining 3 envelopes.
Therefore:

Probability that only one letter will be put into envelop with correct address=
4x(1/4)x(2/3)x(1/2)x(1/1) = 1/3
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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04 Aug 2012, 12:07
Yes, the total # of possible placements for the three letters is 3! = 6 in general, but in this case when we deal with the probability of these letters not being mapped to the right address, we have to take out the right one and so it becomes 2! = 2/letter.
Also, I did not have to compute the same again for other letters, meaning, P(A alone mapped to the right address, but the rest not) = P(B alone mapped to right address while others are not)........ = 2*2*2/4! = 1/3?
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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04 Aug 2012, 13:38
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raghupara wrote:
Yes, the total # of possible placements for the three letters is 3! = 6 in general, but in this case when we deal with the probability of these letters not being mapped to the right address, we have to take out the right one and so it becomes 2! = 2/letter.
Also, I did not have to compute the same again for other letters, meaning, P(A alone mapped to the right address, but the rest not) = P(B alone mapped to right address while others are not)........ = 2*2*2/4! = 1/3?

(1) Using combinatorics:
To place one letter correctly (say A) - 1 possibility
The second letter to misplace it - 2 possibilities
The third letter to misplace it - 1 possibility
The fourth letter will be surely misplaced - just 1 possibility left
Consider the above scenario 4 times - 4 different possibilities for the correctly placed letter (A, B, C, or D can be placed correctly)
Therefore, the probability for exactly one letter placed correctly is 4 * 2 * 1 * 1/4! = 8/24 = 1/3

(2) Using probabilities:
To place one letter correctly (say A) - 1/4
The second letter to misplace it - 2/3
The third letter to misplace it - 1/2
The fourth letter will be surely misplaced - 1/1
So, the probability of a certain letter to be placed correctly and all the others to be misplaced is
(1/4)*(2/3)*(1/2) *(1/1) = 1/12
Again, consider the above scenario 4 times - it gives 4*(1/12) = 1/3

In your answer there are no probabilities (probabilities are expressed as fractions, between 0 and 1 inclusive).
Using the fundamental formula for probabilities (the number of desired outcomes / the total number of possible outcomes) again, your explanation isn't correct.
1*2*2*2 doesn't match any suitable scenario for placing exactly one later correctly and misplace all the other three.
Pay attention to getting the correct answer by using the correct reasoning.
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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05 Aug 2012, 10:19
In a previous thread, IanStewart posted a solution to this question:

"One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3. "

I would like to point out that, although the final result is correct (1/3), the explanation is not correct.
"One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left." This assumption is nowhere reflected in the chain of computations.
In the above computations, 2/3 is what is called a conditional probability, meaning that under the assumption that a certain letter is placed correctly, then the probability of placing one of the remaining 3 letters incorrectly is 2/3, then the probability of placing one of the remaining two letters incorrectly after two were already placed, one correctly and one incorrectly, etc.

The correct chain should be: 1/4 for the probability of a certain letter to be placed correctly, 2/3 for one of the remaining three placed incorrectly, 1/2 for one of the remaining two to be misplaced, and 1/1 for the last one which will be surely misplaced. This gives (1/4)*(2/3)*(1/2)*(1/1)=1/12 for the probability of exactly one particular letter to be placed correctly.
Since there are four possibilities to chose the one letter placed correctly, the required probability is 4*(1/12) = 1/3.

There is a collection of questions with all the possible scenarios for placing the 4 letters:
letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letters%20envelopes

Can we use a similar reasoning used by IanStewart (for exactly 1 correctly placed letter) to find the probability of placing exactly 2 letters correctly?
The reasoning would go like this: consider two letters placed correctly, doesn't matter which ones, then the probability of placing one of the remaining letters incorrectly is 1/2, and then the fourth letter incorrectly is 1/1. Can we conclude that 1/2 is the probability of placing exactly two letters correctly?
The correct answer is 1/4. So, what's wrong with this reasoning?

I claim, we have to take into account the probability of placing a given pair of letters correctly. This probability is 1/12, since the probability of one of the letters to be placed correctly is 1/4, then the other one to be placed correctly is 1/3, which gives the (1/4)*(1/3) = 1/12 (or, either only the two letters are placed correctly or all four are, which means a probability of 2/24 = 1/12).
Then, the probability of one of the remaining two letters to be misplaced is 1/2, and the last one is for sure going to be misplaced (probability 1/1).
Since there are 4C2=(4*3)/2 = 6 possibilities to chose a pair of letters from the given 4 letters, the probability of placing exactly any two letters correctly is given by (1/12) * (1/2) * 6 = 1/4, which is the correct answer.

My point is: all the assumptions should be reflected in the computations, and first of all, we should make all the necessary and correct assumptions, regardless whether we are dealing directly with probabilities or counting number of possibilities. Easier said than done...but we have to try :O)
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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Updated on: 18 Sep 2012, 00:59
clearmountain wrote:
Max 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? answer is 1/3 but very confused.

4 is not a big number, so you can deal with the computations directly.

The total number of possibilities to place the 4 letters in 4 envelopes is 4!.

Any one of the 4 letters can be placed correctly, so 4 possibilities.
After 1 letter is placed in its correct envelope, think of how many ways are there to place the remaining 3 letters such that neither one is in its correct envelope.
3! is the number of possibilities to arrange the three letters, say A,B,C in their envelopes.
Out of these, one possibility when each letter is in its correct envelope - 1
Then, there are 3 more possibilities when one letter is correctly placed and the other two are switched between them - 3
Therefore, a total of 3! - 4 possibilities to have exactly one letter placed correctly.

Required probability is 4(3! - 4)/4! = 4*2/(2*3*4) = 1/3.
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Originally posted by EvaJager on 18 Sep 2012, 00:04.
Last edited by EvaJager on 18 Sep 2012, 00:59, edited 1 time in total.
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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17 Mar 2013, 10:19
1
1
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

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

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18 Mar 2013, 07:55
4
2
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

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

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

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18 Mar 2013, 21:33
6
3
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

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

Re: Tanya prepared 4 different letters to be sent to 4 different addresses   [#permalink] 18 Mar 2013, 21:33

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