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

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

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11 Oct 2009, 10:51
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Difficulty:

95% (hard)

Question Stats:

43% (02:12) correct 57% (01:28) wrong based on 754 sessions

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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
[Reveal] Spoiler: OA
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Re: 4 letters & 4 envelopes [#permalink]

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11 Oct 2009, 11:12
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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

[Reveal] Spoiler: OA
D) 1/3

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

P(C=1)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html
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Re: 4 letters & 4 envelopes [#permalink]

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10 Nov 2009, 09:04
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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
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Re: 4 letters & 4 envelopes [#permalink]

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10 Nov 2009, 09:50
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You can also check the topic below, with almost all possible scenarios for this problem:

letter-arrangements-understanding-probability-and-combinats-84912.html?highlight=Tanya
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Re: 4 letters & 4 envelopes [#permalink]

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17 Mar 2013, 09:19
Bunuel wrote:
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

[Reveal] Spoiler: OA
D) 1/3

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

P(C=1)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html

Hi,

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!
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Re: 4 letters & 4 envelopes [#permalink]

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18 Mar 2013, 06:55
alex1233 wrote:
Bunuel wrote:
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

[Reveal] Spoiler: OA
D) 1/3

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

P(C=1)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html

Hi,

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,

4 * 1/12 = 1/3

Hence the answer
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Re: 4 letters & 4 envelopes [#permalink]

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18 Mar 2013, 20:33
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alex1233 wrote:
Bunuel wrote:
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

[Reveal] Spoiler: OA
D) 1/3

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

P(C=1)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html

Hi,

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!

Check out all three letter and four letter scenarios here:
http://www.veritasprep.com/blog/2011/12 ... envelopes/
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Re: 4 letters & 4 envelopes [#permalink]

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30 Nov 2013, 02:16
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: 4 letters & 4 envelopes [#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: 4 letters & 4 envelopes [#permalink]

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16 Jun 2014, 09:19
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

P(C=1)=8/24=1/3

To check all other possible scenarios check: letter-arrangements-understanding-probability-and-combinats-84912.html[/quote]

Hi,

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]

Check out all three letter and four letter scenarios here:
http://www.veritasprep.com/blog/2011/12 ... envelopes/[/quote]

THANKS A LOT ! THE EXPLANATION IS NOT ONLY CONVINCING BUT ALSO EASY TO GRASP.
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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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

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

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13 Aug 2016, 00:27
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Re: Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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06 Nov 2016, 14:46
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

I think this method may have flaws. What if the second letter is in the right envelope? In this case, those probabilities would be:
Probability that first letter in the wrong envelope= 3/4
Probability that second letter in right envelope = 1/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1
3/4*1/3*1/2*1=1/8

I am confused. Please explain. Thank you.
Re: Tanya prepared 4 different letters to be sent to 4 different   [#permalink] 06 Nov 2016, 14:46
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