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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] New post 26 Oct 2005, 21:18
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A
B
C
D
E

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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
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 [#permalink] New post 26 Oct 2005, 22:27
Ans: C

Number of possible outcomes = 4 * 4 (no of letters * envelopes)
Number of favourable outcome = 4

Prob = Number of favourable outcome/Number of possible outcomes
= 4/4*4 = 1/4

Pls correct me if I am wrong.
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 [#permalink] New post 26 Oct 2005, 23:14
Sorry the OA is (d) or 1/3. Actually this is tricky. The only I figured it out was by writing out all the combinations and....
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 [#permalink] New post 27 Oct 2005, 00:07
There are 4 letters = A, B, C, D
There are 4 addresses = W, X, Y, Z
The correct match = (A, W), (B, X), (C, Y), (D, Z)

Let's calculate that only A goes into the right envelop and all the other 3 go into wrong envelops.
=> 1/4 * 2/3 * 1/2 * 1 = 1/12

From here, we can notice the following;

The possibility that only A goes into the right envelop = 1/12
The possibility that only B goes into the right envelop = 1/12
The possibility that only C goes into the right envelop = 1/12
The possibility that only D goes into the right envelop = 1/12
----------------------------------------------------------------------
Total possibility that only one of the four goes into the right envelop = 1/12 * 4 = 1/3
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 [#permalink] New post 27 Oct 2005, 03:53
Total ways to arrange letters is 4!, because we have 4 letters and 4 envelopes.

Now we have to find the ways, such as 1 letter is placed correct and ALL other letters must be misplaced.
If we put 1 letter in the correct envelope... then there are two ways, of misplacing all other letters incorrectly. two ways, because of the three letters, there are two letters, which can be placed wrongly into on envelope. Placing on of the two in the first wrong envelope, eliminates any further combinations.

However we can place the correct letter in the correct envelope 4 times.

Favorable/Total = 4*2 / 4! = 4*2 / 4*3*2*1 = 1/3
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 [#permalink] New post 27 Oct 2005, 04:38
this is an extremely tricky one..... i don't think on the exam day I will get this one right.
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 [#permalink] New post 27 Oct 2005, 05:24
nakib77 wrote:
this is an extremely tricky one..... i don't think on the exam day I will get this one right.


You will. :wink:
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 [#permalink] New post 27 Oct 2005, 05:53
nero44 wrote:
Total ways to arrange letters is 4!, because we have 4 letters and 4 envelopes.

Now we have to find the ways, such as 1 letter is placed correct and ALL other letters must be misplaced.
If we put 1 letter in the correct envelope... then there are two ways, of misplacing all other letters incorrectly. two ways, because of the three letters, there are two letters, which can be placed wrongly into on envelope. Placing on of the two in the first wrong envelope, eliminates any further combinations.

However we can place the correct letter in the correct envelope 4 times.

Favorable/Total = 4*2 / 4! = 4*2 / 4*3*2*1 = 1/3


Very well explained!! :good
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  [#permalink] 27 Oct 2005, 05:53
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