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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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27 Oct 2008, 10:54
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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

Thanks

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Joined: 31 Jul 2008
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27 Oct 2008, 12:24
d

no. of ways for the 4 letters to go in four different envelopes (total number of events) = 4! = 24

Favorable events = 2*2*2 (the 3 letters each can go in only 2 of the 3 envelopes coz 1 of the 3 has the right address for it.)

p = 8/24 = 1/3

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VP
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27 Oct 2008, 14:21
4 letters in 4! ways

one correct 3 incorrect

1 correct letter can be picked in 4 ways

3 letters 3 spaces such that none is in correct position.

First of the 3 can be placed wrongly in 2 ways. remaining 2 in one way So total is 4 X 2 =8

Probability = 8/24 = 1/3

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27 Oct 2008, 15:46
stallone wrote:
d

no. of ways for the 4 letters to go in four different envelopes (total number of events) = 4! = 24

Favorable events = 2*2*2 (the 3 letters each can go in only 2 of the 3 envelopes coz 1 of the 3 has the right address for it.)

p = 8/24 = 1/3

Can you explain why there are 24 combinations of 4 envelops and 4 letters?
Thanks

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VP
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27 Oct 2008, 17:12
nganle08 wrote:

Can you explain why there are 24 combinations of 4 envelops and 4 letters?
Thanks

first letter can be placed into any of the 4 envelopes - 4 ways, similarly 2nd letter can be placed into 3 envelopes in 3 ways, 3rd letter in 2 ways and 4th letter in 1 way. So it's 4X3X2X1. In other words 4!

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27 Oct 2008, 21:27
icandy wrote:
nganle08 wrote:

Can you explain why there are 24 combinations of 4 envelops and 4 letters?
Thanks

first letter can be placed into any of the 4 envelopes - 4 ways, similarly 2nd letter can be placed into 3 envelopes in 3 ways, 3rd letter in 2 ways and 4th letter in 1 way. So it's 4X3X2X1. In other words 4!

Thanks, icandy. I understand it very clearly now.

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29 Oct 2008, 00:48
1
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My approach:

Probability to put the letter in correct envelope = 1/4
If the first letter is put in the correct envelope, then rest three letters will be put into incorrect envelopes. and probability for this is 1/4*2/3*1/2*1 = 1/12

There are four such possibilities. Hence, total probability = 1/12*4 = 1/3.

Kudos [?]: 279 [1], given: 0

Re: PS: Probability   [#permalink] 29 Oct 2008, 00:48
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