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

Hello there, good day

I'd love to hear your thoughts on this problem here below, I encountered it off one of my CAT practice test from the official prep

I'll write out the question as I remember it, so accurate meaning but not word for word, I also don't remember all the answer choices so I'll leave those out. Here goes

"You have 4 different letters to be sent to different addresses and 4 envelops with the different addresses printed on them. What is the probability that one letter is send to the wrong address"

I have confusion on several parts, but looking at it step by step, this is my logic

1- Probability is the number of outcomes which give us our criteria over the total number of outcomes. The total number of outcomes is 4! since this is essentially an ordered list, so the next question is, how many of those 4! outcomes satisfy the criteria of one letter put in the wrong envelope

2- If one letter is put in the wrong envelope, then at least another letter is in the wrong envelop as well. So this is equivalent to : Wrong/Wrong/*/*

This is where I start to get confused a bit, should we just simply count all the combinations in which there are at least two wrongs ? or should we pin/glue those two wrong and count only for the remaining 2! choices ? And if we do that, do we further need to multiply that result by 4 to account for circular list ?

I think the major points of confusion for me are :
a- I don't recall the question saying exactly one wrong, and I think that's impossible anyways, given that there are 4 distinct letters and 4 envelopes
b- I am quite confused about the "circular" logic list, if anything i'd really appreciate it if you could answer even just this

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

A more cumbersome but relatively straight fwd approach for probability-sick people like me

General counting method should work fine here (since we're dealing with only 4 envelopes)

ABCD
ABDC
*ADBC
ADCB
ACBD
*ACDB

BACD
BADC
BDAC
*BDCA
*BCAD
BCDA

*CABD
CADB
CDAB
CDBA
CBAD
*CBDA

*DACB
DABC
DBAC
*DBCA
DCBA
DCAB

Only arrangements with * marked against them have 1 letter correctly places in its corresponding address, hence probability = 8/24 = 1/3. Hence (D).

Sorry if I am replying to a very old post, But I know a formula which can be used for the above problem.

The above problem comes under a category called DERANGEMENT: Where you basically asked to match objects with WRONG pairs. Like putting a letter into incorrect envolope

Formula is pretty straight forward:

Lets consider this question, where you are asked to put only 1 letter in the correct envelope that implies that you need to put the other THREE letters in incorrect envelopes:

d(n) = n!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}.......\frac{(-1)^n}{n!})

D(n) = \frac{d(n)}{n!}

in this case we need to calculate for n=3

d(3) = 3! * (\frac{1}{2!}-\frac{1}{3!}) = 2
D(3) = \frac{d(3)}{3!} = \frac{1}{3}

Plese note that the only condition here is n >= 2, As you know that probality of putting only 1 letter in incorrect envelope is ZERO as it is not possible situation

Check this for all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html

Hope it helps.
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My 2 cents:

For 1 letter to go in right envelope - probability = 1/4
For 2nd letter to go in wrong envelope - probability = 2/3
For 3rd letter to go in wrong envelope - probability = 1/2
For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done:
C - Letter going into correct envelope
W - Letter going into wrong envelope

No#1. C - W - W - W
No#2. W - C - W - W
No#3. W - W - C - W
No#4. W - W - W - C

There are 4 ways, it can be done

So, probability of 1 letter going in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3

You are presenting your solution for exactly one correctly placed letter?
I guess, you wanted to find in how many ways B,C and D can be placed so that none of them is in the right envelope.
This cannot give 2*2*2 = 8, because the total number of possible placements for the three letters is 3! = 6 < 8.

In addition, what about the cases when only B is placed correctly? Only C? Only D?
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

(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.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

In a previous thread, IanStewart posted a solution to this question:
(tanya-prepared-4-different-letters-to-4-different-addresses-83683.html)

"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)
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.