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

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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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!]

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 -



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 -



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 -



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 -



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 -



Now I think it is easy for you to go around............
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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

@
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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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

For more info on the concept of derangements please refer to the Experts' Global concept video:

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
- adbc
- 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
- bcad
- cabd
There are 8 such outcomes.
-------------------

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

Answer: D

Cheers,
Brent

One letter which goes in correct envelope can be chosen in 4C1 ways = 4 ways

Second letter can choose wrong envelope in 2 ways (except its correct envelope)

Third letter can accept only one envelope in order to ensure that 4th also goes in wrong envelope

i.e. Total Favourable Ways = 4C1*2*1*1 = 8 ways

Total ways to put 4 letters in 4 envelopes = 4! = 24

Probability = 8/24 = 1/3

Answer: Option D

Solution:

Let ABCD be the correct order of the 4 letters in the 4 envelopes. For example, the arrangement ABDC means exactly 2 letters are put correctly in the envelopes and the arrangement ACDB means exactly 1 letter is put correctly in the envelopes. We know that there are 4! = 24 such arrangements. Now let’s list all the arrangements such that letter A is put correctly into its envelope (the alphabets in bold are the letters that are put correctly into their envelopes):

ABCD, ABDC, ACBD, ACDB, ADBC, and ADCB

Notice that in the 6 arrangements above, only 2 (the underlined ones) have only letter A (i.e., exactly 1 letter) put correctly in its envelope. Using the same arguments, we can say that if we list all the arrangements such that letter B is put correctly into its envelope, only 2 (out of 6) will have only letter B put correctly in its envelope. We can say the same thing for letters C and D.
Therefore, the number of arrangements where exactly 1 letter is put correctly in its envelope is 2 x 4 = 8. Since there are a total of 24 arrangements, the probability that exactly 1 letter is put correctly in its envelope is 8/24 = 1/3.

Answer: D
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If someone is having a hard time understanding this solution, refer this worked out example.

Let the letters be 1,2,3,4 and address envelopes be 1a,2a,3a,4a

Let letter 1 be put in its correct envelope 1a
So,
2 can be put in 3a or 4a
3 can be put in 2a or 4a
4 can be put in 2a or 3a

1-1a
2-4a
3-2a
4-3a

1-1a
2-3a
3-4a
4-2a

Thus for 1-1a to be correct and others to be wrong there are 2 cases.
Now, there are 3 other such letters, i.e 2-2a will be correctly done and others will be wrong, 3-3a others wrong and 4-4a others wrong. Total 4 letters.
Therefore, there are 4 (letters) * 2 (cases when one letter is in correct envelope and others arent) = 8 such cases

Total number of ways letter can be put is 4! = 24

Reqd probaility = 8/24 = 1/3

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?

Assume there are 4 letters A,B,C,D and 4 corresponding envelops for each letter.

Total no ways you can arrange 4 letters in 4 envelopes = 4! = 24 ways

No of ways where only one letter will be put into the envelope with its correct address and other letters will be in wrong envelops = 2+2+2+2 = 8 ways

Please refer the below table to get an idea about how we got 8 ways




The probability that only one letter will be put into the envelope with its correct address = Favourable outcome / Total outcome = 8/24 = 1/3

Option D is the answer.

Thanks,
Clifin J Francis,
GMAT SME
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Bonjour !!

fortunately, this method did strike in the exam and i got the right ans. However, i have been thinking about the below approach since i did review.
Lets consider
case -1: 1st goes to right and the others go to wrong.. so, prob = 1/4*2/3*1/2*1/1
case -2 1st in wrong env and 2nd in correct and then 3rd and 4th in wrong ; prob = 3/4* 1/3*1/2*1/1
case -3 1st in wrong , 2nd in wrong , 3rd in right, 4th in wrong ; prob = 3/4*2/3*1/2*1/1
case-4 first 3 in wrong and last in correct ; prob = 3/4*2/3*1/2*1/1

total prob = add all 4 cases = 17/24.

i am keen to know - where is my reasoning wrong? IMO, with this approach i am not inserting simultaneously and trying to pick and insert one by one. Is that what it is??
Please help .

Case 1 is correct. Cases 2-4 are not.

Case 2: Using 3/4 for A in any of the other envelopes includes the possibility of putting in B, which will then render probability of B going into B equal to 0 and the subsequent chain of probabilities 0 since 0 is being multiplied.

So, need to look at probability of A going into C which is then 1/4.

Then probability of B going to B is then 1/3.

C then has to go to D with probability of 1/2 in order for D to A.

Probability of this is 1/4 * 1/3 * 1/2 or 1/24.

Then need to look at A going to D, and then follow the same approach, which gives another 1/24.

So probability of B only being properly inserted is 1/12.

C and D follow same approach as above.

So there are 4 sets of 1/12 probability, equaling 1/3 in total.

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Hi could someone help?

TOtally understood for the first case where it is Correct Wrong Wrong Wrong.

I actually wrote out all the different cases and summed it up could someone tell me why my approach is wrong?

E.g.
WCWW = 3/4*1/3*1/2 = 1/8 ;

WWCW = 3/4*2/3*1/2 = 1/4 ;

WWWC = 3/4*2/3*1/2 = 1/4

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Total ways of derrangement of 3 letters= 3!(1/0!-1/1!+1/2!-1/3!)
Total ways of derrangement of 3 letters with correct placement of 4th letter= 4C1*3!(1/0!-1/1!+1/2!-1/3!)
=8 ways

P=8/4!=8/24=1/3

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You can also solve this problem with derangements i.e. no of ways so that the envelope does not go into the right envelope.