letters and envelopes - probability

[DavoodBeater]

Please explain. 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?

OA: [spoiler]1/3[/spoiler]

[logitech]

A B C D
1 2 3 4

Let's say only A1 is matched correctly.

For 2 can choose only C or D
For 3 can choose only D or B
For 4 can choose only B or C

So actually for every letter that is in its correct address(A1) you can have two different ways to distribute the other three to ALL incorrect addresses.

Since we have 4 letters

4x2= 8 Desired Scenarios

4 letter can be sorted as 4! ways

8/4! = 1/3

[sachinkr]

Let's Assume 4 letters are labeled a,b,c,d and envelope are labeled A,B,C,D.

Total number of ways 4 letters can be put in 4 envelopes = 4*3*2*1 = 24

Number of ways only 1 letter is in correct envelope

In case label 'a' letter is put in label A envelope. the number of cases are :
1: A (a), B (c) , C (d), D (b)
2: A (a), B (d) , C (b), D (c)

Similarly b,c,d letters can be placed in their respective envelope. So in all there are 4 * 2 = 8.

So the probability is 8/24 => 1/3.

[DavoodBeater]

Thanks a lot for your answers.

[amitabhprasad]

Four letter say labeled as A,B,C,D.
P(A) is in the right envelop = 1/4
P(B) is in the right envelop = 1/3
==> P(B) is not in the right envelop = 2/3
P(C) is int he right envelop = 1/2
==> P(C) is not in the right envelop = 1-1/2 = 1/2
==>P(D) not in the right envelop = 1
Thus probability of only getting "A" in the right envelop =
P(A) && P(!B) && P(!C) && P(!D) = P(A)*P(!B)*P(!C)*P(!D) = 1/4*2/3*1/2*1 = 1/12
Similarly probability "B"of P(B) = 1/12 ; P(C) = 1/12;P(D) = 1/12

We have 4 envelop chance of any one of them is wrong =
P(A or B or C or D) = P(A)+P(B)+P(C)+P(D) = 1/12+1/12+1/12+1/12
=4/12 = 1/3

[DavoodBeater]

I did the same.
but you know, the first success is not always the first one. So if you faced a failure at first, the chance of success would be 1/3. so;
3/4 * 1/3 * 1/2 * 1
but this way will not lead us to the same answer and I dont know why.
the whole answer which I did:

1/4 * 2/3 * 1/2 * 1 : first success
3/4 * 1/3 * 1/2 * 1 : second success
3/4 * 2/3 * 1/2 * 1 : third success
3/4 * 2/3 * 1/2 * 1 : last success
the sum is 17/24 not 8/24