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

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Joined: 09 Jul 2012
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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15 Sep 2013, 11:23
6
1
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.
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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30 Nov 2013, 02:16
1
Wayxi wrote:
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

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

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30 Nov 2013, 03:19
archit wrote:
Wayxi wrote:
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

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

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10 Aug 2014, 22:25
2
1
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!]
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Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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23 Apr 2016, 01:16
The trick is " Understanding the basics of concept of Counting and Probability "

Ok , I will provide a diagrammatic representation to elucidate the issue...

You can arrange the first letter in the correct envelop as -

Attachment:

L1.PNG [ 5.24 KiB | Viewed 2035 times ]

Here u can arrange the first letter in 4 ways (Any envelop E1 , E2 , E3 , E4) , so you have 4 choices for placing the letter.

However keep in mind that only one choice L1 - E1 is the correct choice , rest 3 choices are wrong.

You can arrange the second letter in the correct envelop as -

Attachment:

L2.PNG [ 4.38 KiB | Viewed 2035 times ]

Here u can arrange the second letter in 3 ways (Any envelop E2 , E3 , E4) , so you have 3 choices for placing the letter.

However keep in mind that only one choice L2 - E2 is the correct choice , rest 2 choices are wrong.

You can arrange the third letter in the correct envelop as -

Attachment:

L3.PNG [ 2.86 KiB | Viewed 2034 times ]

Here u can arrange the second letter in 2 ways (Any envelop E3 , E4) , so you have 2 choices for placing the letter.

However keep in mind that only one choice L3 - E3 is the correct choice , and the other choice is wrong.

You can arrange the forth letter in the correct envelop as -

Attachment:

L4.PNG [ 1.94 KiB | Viewed 2026 times ]

Here u can arrange the fourth letter in only 1 way (E4)

Now Follow my colour coding very carefully

Possible arrangement of the Letters in the envelop (Irrespective of correct/wrong) will be 4 * 3 * 2 * 1 = 24

Read the question stem very carefully it requires us to find -

Quote:
the probability that only 1 letter will be put into the envelope with its correct address

Now I think it is easy for you to go around............
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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24 Apr 2017, 17:41
5
Wayxi wrote:
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

For anyone wondering about this solution, it is actually correct only by coincidence. This method will fail when the number of letters increased to 5, and is therefore an incorrect formula.

"There are issues with this calculation. it happens to hit upon the correct value at the end, but that's a total coincidence.

i agree with the first two probabilities: the probability that letter a goes into envelope a is indeed 1/4, and the probability if that happens that the letter b goes into an envelope other than b is 2/3.
however, it's downhill from there: if letter b actually went into envelope c, then the probability of letter c not going into envelope c is 1. the probability is only 1/2 (as you've stated) if letter b winds up in envelope d.
similarly, the final probability is either 0 or 1, depending on whether the last envelope remaining is envelope d or not. if letter b goes in envelope c and letter c goes in envelope b (fulfilling all of your conditions), then letter d is stuck going into envelope d, making that last probability 0.

so, if you're going to go this route, you're stuck with doing the following:
* first 2 steps = same as you have them now
* 3rd step = 2 branches of a probability tree, depending on whether envelope c is still available (vs. whether it was used for letter b)
* 4th step = 2 branches off EACH of those prior 2 branches, depending on whether envelope d is still available (vs. whether it was used for letter b or c)"

If you would like to read further, see Manhattan Prep's forum post on this question.

This is a quote from Ron Purewal
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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31 May 2017, 02:55
2
@
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

1. Total number of arrangements is 24
2. List out the arrangements partly, if letter 1 is in the first envelope
They are 1234, 1243, 1324,1342, 1423, 1432
3. Out of the six above cases, 2 cases satisfy the condition. It will be the same for the other letters in the first envelope
4. So the probability is 8/24 which is 1/3.
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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17 Jul 2019, 20:13
One can simply solve these kind of questions by the concept of Derangements.

Derangement is given by the formula:

$$D_n$$ = $${1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - ..... (-1)^n \frac{1}{n!}}$$

So in this case, we have to derange the 3 other cards as 1 card is already in its correct envelope.

So $$D_3$$ should be calculated.

$$D_3$$ = $${1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}}$$

$$D_3$$ = $$\frac{1}{2} - \frac{1}{6}$$

$$D_3$$ = $$\frac{1}{3}$$

OPTION: D

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

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29 Dec 2019, 09:52
Top Contributor
tealeaflin 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

Let's solve this question using counting methods.

So, P(exactly one letter with correct address) = (number of outcomes in which one letter has correct address)/(total number of outcomes)

Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses.
So, let's list the letters in terms of the order in which they are delivered to addresses A, B, C, and D.
So, for example, the outcome abcd would represent all letters going to their intended addresses.
Likewise, cabd represent letter d going to its intended address, but the other letters not going to their intended addresses.

-------------------
total number of outcomes
The TOTAL number of outcomes = the number of different ways we can arrange a, b, c, and d

RULE: We can arrange n different objects in n! ways.
So, we can arrange the four letters in 4! ways = 24 ways
-------------------

number of outcomes in which one letter has correct address
Now let's list all possible outcomes in which exactly ONE letter goes to its intended addresses:
- acbd
- cbda
- dbac
Aside: At this point you might recognize that there are two outcomes for each arrangement in which one letter goes to its intended address
- dacb
- bdca
- cabd
There are 8 such outcomes.
-------------------

So, P(exactly one letter with correct address) = 8/24 = 1/3

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

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19 Feb 2020, 05:46
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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I used the same approach on my first attempt but there seems to be some confusion during my revision.

4 (# of ways to arrange the 4 letters) * P(getting 1 letter right) -> we are assuming that all arrangements have the same probability, right?

However, it does not seems the case as shown below.

RWWW: 1/4 * 2/3 * 1/2 * 1/1 = 1/12
WRWW: 3/4 * 1/3 * 1/2 * 1/1 = 1/8
WWRW: 3/4 * 2/3 * 1/2 * 1/1 = 1/4
WWWR: 3/4 * 2/3 * 1/2 * 1/1 = 1/4

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

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19 Feb 2020, 06:19
Required probability
=4[1×2×1×1]÷[4×3×2×1]=1/3

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

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19 Feb 2020, 09:42
Assume four letters L1, L2, L3 and L4 are to be send to their corresponding address A1, A2, A3 and A4.
Now one letter can be send to it's proper address in. any of four ways namely
L1-A1
L2-A2
L3-A3
L4-A4
And three other letters can. be sent to their wrong address in
2×1×1=2 ways
Therefore required probability
= (4×2)/(4×3×2×1) =1/3
Hence correct answer is option (D).
Re: Tanya prepared 4 different letters to be sent to 4 different addresses   [#permalink] 19 Feb 2020, 09:42

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