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

It is currently 30 Aug 2014, 00:27

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

I know this is not typically a hard one but I have a

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 2 [0], given: 0

GMAT Tests User
I know this is not typically a hard one but I have a [#permalink] New post 10 Nov 2005, 19:39
I know this is not typically a hard one but I have a question about it:

A fair die has sides labeled w/ 1, 2, 3, 4, 5, 6. If the die is rolled 4 times, what is the probability that on at least 1 roll, the die will show 6?


Now, I'd like to know why you HAVE to solve by looking at it from the probability of NOT getting a six. In other words...why can't you add 1/6 4 times? When is it key to go with probability of not getting what you are trying to get, rather than solving for the probability of getting it?

Many thanks
Director
Director
avatar
Joined: 27 Jun 2005
Posts: 517
Location: MS
Followers: 2

Kudos [?]: 10 [0], given: 0

GMAT Tests User
Re: Rolling a die [#permalink] New post 10 Nov 2005, 23:04
Jennif102 wrote:
I know this is not typically a hard one but I have a question about it:

A fair die has sides labeled w/ 1, 2, 3, 4, 5, 6. If the die is rolled 4 times, what is the probability that on at least 1 roll, the die will show 6?


Now, I'd like to know why you HAVE to solve by looking at it from the probability of NOT getting a six. In other words...why can't you add 1/6 4 times? When is it key to go with probability of not getting what you are trying to get, rather than solving for the probability of getting it?

Many thanks


Jennif102,

the die rolling is a independent event so the total probability will be the multipication of probability of each event.

and the case you are refering where using 1/6 as the probabiltiy of getting 6 .....1/6*1/6*1/6*1/6 .. is the probabiltiy of getting 6 in each event not getting at least one six.

but in the question the probability of at least one 6 is asked .....so in 4 rolls you may have 1 six or 2 six or 3 six ...or all 4 six ......so the best way is to 1 - probability of None= probability of at least one.....otherwise you have to consider the cases all the case

1six3Nonesix+2six2Nonesix...and so on


hope it helps....
Manager
Manager
avatar
Joined: 12 Mar 2005
Posts: 65
Location: Milton Keynes UK
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 11 Nov 2005, 06:30
Cool explanation Jonny009.
_________________

The ability to focus attention on important things is a defining characteristic of intelligence.
--by Robert J. Shiller

SVP
SVP
User avatar
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 40 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Nov 2005, 07:12
i will do it as

1/6^4 +2/6^4+ 3/6^4+ 4/6^4

= 1/6^4 ( 1+2+3+4)

= 10/6^4
_________________

hey ya......

Manager
Manager
avatar
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 2 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Nov 2005, 16:48
thanks! Very helpful
Senior Manager
Senior Manager
avatar
Joined: 07 Jul 2005
Posts: 405
Followers: 3

Kudos [?]: 11 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Nov 2005, 20:10
Here you go:

P(xxxx) + P(6xxx) + P(66xx) + P(666x) + P(6666) = 1

P(6xxx) + P(66xx) + P(666x) + P(6666) = 1 - P(xxxx)

Since each occurance is an independent event.

P(xxxx) = P(x1) * P(x2) * P(x3) * P(x4)

The probability of not getting a 6 is 5/6

P(xxxx) = (5/6)^4

Hence,

P(6xxx) + P(66xx) + P(666x) + P(6666) = 1 - (5/6)^4
  [#permalink] 11 Nov 2005, 20:10
    Similar topics Author Replies Last post
Similar
Topics:
I know for the other sections it's easy to hard. Is that h3nG 1 27 Jul 2011, 17:31
One more hard one before I go and be merry! I will post lfox2 3 24 Dec 2006, 08:12
I have been studying really hard rbcola 10 16 Nov 2006, 13:55
Here another one... I am just having hard time figuring this lan583 7 02 Aug 2006, 04:36
this is from one of the 'hard math' princeton pdf's and i user 7 05 Sep 2004, 21:00
Display posts from previous: Sort by

I know this is not typically a hard one but I have a

  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®.