Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with

[Gmat_mission]

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

Answer: D

Source: GMAT Prep

[Brent@GMATPrepNow]

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

Answer: D

Source: GMAT Prep

Let's solve this question using counting methods.

So, P(exactly one letter with correct address) = (number of outcomes in which one letter has correct address)/(total number of outcomes)

Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses.
So, let's list the letters in terms of the order in which they are delivered to addresses A, B, C, and D.
So, for example, the outcome abcd would represent all letters going to their intended addresses.
Likewise, cabd represent letter d going to its intended address, but the other letters not going to their intended addresses.

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total number of outcomes
The TOTAL number of outcomes = the number of different ways we can arrange a, b, c, and d

RULE: We can arrange n different objects in n! ways.
So, we can arrange the four letters in 4! ways = 24 ways
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number of outcomes in which one letter has correct address
Now let's list all possible outcomes in which exactly ONE letter goes to its intended addresses:
- acbd
- adbc
- cbda
- dbac
Aside: At this point you might recognize that there are two outcomes for each arrangement in which one letter goes to its intended address
- dacb
- bdca
- bcad
- cabd
There are 8 such outcomes.
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So, P(exactly one letter with correct address) = 8/24 = 1/3

Answer: D

Cheers,
Brent