Find all School-related info fast with the new School-Specific MBA Forum

It is currently 24 Apr 2014, 19:28

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

As part of a game, four people each must secretly choose an

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
Joined: 03 Sep 2005
Posts: 136
Followers: 2

Kudos [?]: 0 [0], given: 0

As part of a game, four people each must secretly choose an [#permalink] New post 16 Sep 2005, 09:43
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A) 9%
B) 12%
C) 16%
D) 20%
E) 25%

I tried to solve this by doing (1/4)(1/3)(1/2)(1), but that didn't work. Can someone explain why that is incorrect? My thought process was that there is a 1/4 chance the first person will pick a certain number, then since there are only 3 numbers left, there is a 1/3 chance, etc... Now I'm all confused. Help?
_________________

Life is not measured by the number of breaths you take, but by the number of moments that take your breath away.

Manager
Manager
Joined: 03 Sep 2005
Posts: 136
Followers: 2

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 Sep 2005, 09:52
need700+, can you please explain your logic?

need700+ wrote:
i get A

4/4*3/4*2/4*1/4=24/256=9.385%

_________________

Life is not measured by the number of breaths you take, but by the number of moments that take your breath away.

Senior Manager
Senior Manager
Joined: 27 Aug 2005
Posts: 332
Location: Montreal, Canada
Followers: 2

Kudos [?]: 3 [0], given: 0

GMAT Tests User
 [#permalink] New post 16 Sep 2005, 10:12
This may be a bit less mathematical but I also got A.

Probability = desired outcomes / total outcomes.

Total outcomes = 4^4 = 256.

Desired outcomes = 6*4 = 24

To get to the 24 desired outcomes, I simply went about it methodically. If the first person drew a 1, how many ways are there for the other three to avoid repeats. There are 6 ways: 1234, 1243, 1324, 1342, 1423, and 1432. The same thing repeats when the first person chooses 2, 3 or 4. So there are 6x4 or 24 desired outcomes.

24/256 = 3/32 = ~9%
VP
VP
Joined: 22 Aug 2005
Posts: 1128
Location: CA
Followers: 1

Kudos [?]: 15 [0], given: 0

GMAT Tests User
Re: PS: Permutation Game [#permalink] New post 16 Sep 2005, 10:27
trulyblessed wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A) 9%
B) 12%
C) 16%
D) 20%
E) 25%

I tried to solve this by doing (1/4)(1/3)(1/2)(1), but that didn't work. Can someone explain why that is incorrect? My thought process was that there is a 1/4 chance the first person will pick a certain number, then since there are only 3 numbers left, there is a 1/3 chance, etc... Now I'm all confused. Help?


A.

corrections:
1/4 chance the first person will pick a certain number

you dont have to find probability of first person "picking a number". The question asks about probability of picking DIFFERENT (than the rest 3) numbers. Now, lets solve your ways:

- chance for first person to pick a DIFFERENT number is: 1. As he is first person, his number is always different.

- chance for second person to chose DIFFERENT number = 3/4
- thirs: 2/4
- fourth : 1/4

p = 1 * 3/4 * 1/2 * 1/4 = 3/32 = .0937 or ~9 %
Intern
Intern
Joined: 19 Aug 2005
Posts: 41
Followers: 0

Kudos [?]: 1 [0], given: 0

 [#permalink] New post 16 Sep 2005, 10:57
here is how i did it

we are letting the first person to choose any number among 4 so his chance is 4/4
the second person can choose any 3 numbers among 4.. so his chances is 3/4
the third person can choose any 2 numbers among 4.. so his chance is 2/4
and the last person had only 1 number to choose frm 4 so his chance is 1/4

multiplying all these i get ~9%
Manager
Manager
Joined: 03 Sep 2005
Posts: 136
Followers: 2

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 16 Sep 2005, 11:33
Wow, you guys are soooo awesome. I really appreciate your help!!!
_________________

Life is not measured by the number of breaths you take, but by the number of moments that take your breath away.

  [#permalink] 16 Sep 2005, 11:33
    Similar topics Author Replies Last post
Similar
Topics:
New posts As part of a game, four people each must secretly choose an Seyi 3 02 Dec 2004, 12:42
New posts As part of a game, four people each must secretly choose an MA 4 21 Apr 2005, 22:45
New posts As part of a game, four people each must secretly choose an acfuture 9 11 Jun 2006, 12:20
Popular new posts 15 Experts publish their posts in the topic As part of a game, four people each must secretly choose an tarek99 23 30 Nov 2007, 07:57
Popular new posts 3 Experts publish their posts in the topic As part of a game, four people each must secretly choose an Financier 17 06 Aug 2010, 22:40
Display posts from previous: Sort by

As part of a game, four people each must secretly choose an

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.