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

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

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15 Aug 2008, 04:48
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Tanya prepared 4 different letters to be sent in 4 different envelopes. 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 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

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Director
Joined: 27 May 2008
Posts: 541

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15 Aug 2008, 05:22
tarek99 wrote:
Tanya prepared 4 different letters to be sent in 4 different envelopes. 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 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

lets say letters are numbered 1,2,3,4
total number of combinations = 4!

probability that 1 goes into right envelope
number of ways selecting 1's envelope = 1
number of ways selecting 2's envelope = 2 (1 already gone and cant take 2)
number of ways selecting 3's envelope = 1 (cant take 3)
number of ways selection 4's envelope = 1 ( cant take 4)
fav ways = 2

P = 2/4! = 1/12

but this the probablility of only 1 going into right envelope
probability of any one letter going into right envelope = 4 * 1/12 = 1/3 option D

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15 Aug 2008, 22:27
Total no of combination = 4! = 24
Once one letter is correctly placed, the others three letters can be wrongly placed each in 2 ways (as the third one is the correct address.
So the rest three letters can be wrongly put into 2*2*2 = 8 ways..
So the probability of one letter in right envelope is
= 8/24 = 1/3
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Manager
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Location: Vienna, Austria

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05 Oct 2008, 01:29
guys

why do you only get 2 ways? isn´t it:
A B C D
a b c d
a c d b
a d b c
a b d c

?

thanks for clarification

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Director
Joined: 27 May 2008
Posts: 541

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05 Oct 2008, 08:06
domleon wrote:
guys

why do you only get 2 ways? isn´t it:
A B C D
a b c d ---> this is not a valid selection as both A and B are going in right envelopes
a c d b
a d b c
a b d c ---> this is not a valid selection as both A and B are going in right envelopes
?
thanks for clarification

As per our question, "what is the probability that only one letter will be put into the envelope with its correct address?"

hope that clears the doubt....

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Senior Manager
Joined: 04 Aug 2008
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05 Oct 2008, 10:47
durgesh79 wrote:
domleon wrote:
guys

why do you only get 2 ways? isn´t it:
A B C D
a b c d ---> this is not a valid selection as both A and B are going in right envelopes
a c d b
a d b c
a b d c ---> this is not a valid selection as both A and B are going in right envelopes
?
thanks for clarification

As per our question, "what is the probability that only one letter will be put into the envelope with its correct address?"

hope that clears the doubt....

However, if you are matching 4 with corresponding 4 envelopes

the probability of putting any letter u select (it does not matter which) to any envelope would be 1/4

it cant be 1/12 - it does not matter which letter u select out of 4
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Manager
Joined: 30 Sep 2008
Posts: 111

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06 Oct 2008, 00:59
durgesh79 wrote:
tarek99 wrote:
Tanya prepared 4 different letters to be sent in 4 different envelopes. 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 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

lets say letters are numbered 1,2,3,4
total number of combinations = 4!

probability that 1 goes into right envelope
number of ways selecting 1's envelope = 1
number of ways selecting 2's envelope = 2 (1 already gone and cant take 2)
number of ways selecting 3's envelope = 1 (cant take 3)
number of ways selection 4's envelope = 1 ( cant take 4)
fav ways = 2

P = 2/4! = 1/12

but this the probablility of only 1 going into right envelope
probability of any one letter going into right envelope = 4 * 1/12 = 1/3 option D

D is also my choice

If 1 letter already got 1 correct address, so 3 letters left and these 3 will be put in all incorrect addresses

For the 2nd letter, it has 3 options in which 2 will be incorrect so 2/3
For the 3rd letter, it has 2 options in which 1 will be incorrect so 1/2
The last has no other choice

P = 2/3 * 1/2 = 1/3

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Re: PS: Probability   [#permalink] 06 Oct 2008, 00:59
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