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Tanya prepared 4 different letters to be sent to 4 different [#permalink]
11 Oct 2009, 10:51

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95% (hard)

Question Stats:

40% (02:14) correct
60% (01:32) wrong based on 536 sessions

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?

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?

Could someone please explain a good way to go about solving this please?

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) ADBC(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

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:

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?

Could someone please explain a good way to go about solving this please?

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) ADBC(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

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!

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?

Could someone please explain a good way to go about solving this please?

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) ADBC(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

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,

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?

Could someone please explain a good way to go about solving this please?

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) ADBC(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

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!

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 ?

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. _________________

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) ADBC(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

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![/quote]

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

1

This post received KUDOS

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!]

Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]
11 Aug 2015, 08:00

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