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

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

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New post 08 Sep 2009, 11:00
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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New post 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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New post 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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New post 05 Aug 2012, 10:19
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)
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 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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New post 30 Nov 2013, 04:19
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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New post 17 Jul 2019, 21:13
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

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


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Re: Tanya prepared 4 different letters to be sent to 4 different addresses   [#permalink] 17 Jul 2019, 21:13

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