Need help with probability - following question from original GMAT

[PN123]

Tanya prepared 4 different letters to be sent to 4 different addresses.
For each letter, she prepared an envelop with its correct address. if the 4 letters are to be put into the 4 envelops at random, what is the probability that only one letter will be put into the envelop with its corret address.

a. 1/24
b. 1/8
c. 1/4
d. 1/3
e. 3/8

Thanks in advance since I iwll be testing in about 12 hours.

[popatpandu]

Total possible outcomes = 4 * 4 = 16
Only 1 correct outcome = 4C1 = 4

Hence, answer = 4/16 = 1/4.

IMO, answer = 1/4.

[putneyswope]

People, what am I doing wrong!?

I tried a different approach, starting with what is the probability that they all match.

Imagine four separate steps:

The probability of Letter A going into Envelope A = 1/4
The probability of Letter B going into Envelope B = 1/3
The probability of Letter C going into Envelope C = 1/2
The probability of Letter D going into Envelope D = 1

So the probability that they all match is 1/4x1/3x1/2x1 = 1/24

So the probability that ONLY A matches

The probability of Letter A going into Envelope A = 1/4
The probability of Letter B NOT going into Envelope B = 2/3
The probability of Letter C NOT going into Envelope C = 1/2
The probability of Letter D NOT going into Envelope D = 1

So the probability ONLY A matches is 1/4x2/3x1/2x1 = 2/24 = 1/12

Can anybody point out where I'm going wrong?

thanks!

[popatpandu]

putneyswope,

Why do you think ONLY A has the chance of being "correct"?
It can be anyone from A,B,C or D.

[putneyswope]

aha . . . right . . . thanks popatpandu

[aparajita0908]

it will be a. 1/24
total number of ways in which the three letters can be put in four envelopes is 4*3*2*1
and there is only one correct method so that all are in correct envelope
so 1/24

[ihsuw]

Originally Posted by PN123

Tanya prepared 4 different letters to be sent to 4 different addresses.
For each letter, she prepared an envelop with its correct address. if the 4 letters are to be put into the 4 envelops at random, what is the probability that only one letter will be put into the envelop with its corret address.

a. 1/24
b. 1/8
c. 1/4
d. 1/3
e. 3/8

I think the answer is d. 1/3
Total possible outcomes are 4*3*2*1=24
The probability that only one letter will be put into the envelope with its correct address is 8.
A - a
B - c or B - d
C - d or C - b
D - b or D - c
When only one letter A is put into the correct envelope a, the possible outcomes of this are 2. Also you have to consider the other situation that letters other than A are put into their correct envelopes.
2*4<A.B.C.D>=8

thus, 8/24 = 1/3
the answer is d.

[putneyswope]

yes D

[toll_me]

This is my approach,

Lets take letter A is in correct envelop,
Now we are left with 3 letter and 3 envelopes
Total No of Ways to put 3 letters in 3 envelops = 3x2x1 = 6 ways
Now since letter A is already in Correct envelop the rest of letters B,C,D should be in wrong envelop
i.e. Envelop B can have C or D
Envelop C can have only D
and Envelop D can have only B
= 2x1x1 = 2 ways
That means there are only 2 ways to do this, and total is 6 ways
i.e. 2/6 = 1/3
Even if we multiply this by 4 {considering A,B,C or D can hold correct letter)
then we will get same answer
i.e. (2x4) / (6x4)

Hence Answer is D (1/3)

[anandisonline]

Ya D is rght