It is currently 25 Jun 2017, 08:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Tanya prepared 4 different letters to be sent to 4 different

Author Message
Senior Manager
Joined: 07 Jul 2005
Posts: 402
Tanya prepared 4 different letters to be sent to 4 different [#permalink]

### Show Tags

26 Oct 2005, 22:18
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Senior Manager
Joined: 02 Oct 2005
Posts: 297

### Show Tags

26 Oct 2005, 23:27
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.
Senior Manager
Joined: 07 Jul 2005
Posts: 402

### Show Tags

27 Oct 2005, 00:14
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....
Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea

### Show Tags

27 Oct 2005, 01:07
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
_________________

Auge um Auge, Zahn um Zahn !

Manager
Joined: 04 Oct 2005
Posts: 246

### Show Tags

27 Oct 2005, 04:53
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
SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka

### Show Tags

27 Oct 2005, 05:38
this is an extremely tricky one..... i don't think on the exam day I will get this one right.
_________________

hey ya......

Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea

### Show Tags

27 Oct 2005, 06:24
nakib77 wrote:
this is an extremely tricky one..... i don't think on the exam day I will get this one right.

You will.
_________________

Auge um Auge, Zahn um Zahn !

Director
Joined: 15 Aug 2005
Posts: 796
Location: Singapore

### Show Tags

27 Oct 2005, 06:53
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!!
_________________

Cheers, Rahul.

27 Oct 2005, 06:53
Display posts from previous: Sort by