Find all School-related info fast with the new School-Specific MBA Forum
Learn more about the recent changes... close

It is currently 21 May 2013, 02:11
Customize  |  Hide

Search
Register
New posts
Unanswered

1) One has three fair dice and rolls them together. What is

  Question banks Downloads My Bookmarks Reviews  
  Oldest Best Reply
Search for:
 
Print view
First unread post
Author Message
TAGS:
SVP
SVP
User avatar
Joined: 03 Feb 2003
Posts: 1683
Followers: 4

Kudos [?]: 16 [0], given: 0

1) One has three fair dice and rolls them together. What is [#permalink] New post 17 Aug 2003, 01:38
(1) One has three fair dice and rolls them together. What is the probability of having two same numbers and one different? (for example, 2-2-6, 1-3-3, 6-5-6, and so on)

(2) One has four fair dice and rolls them together. What is the probability of having all the numbers different? (for example, 1-2-3-4, 6-5-4-1, and so on)

(3) One has six fair dice and rolls them together. What is the probability of having all the numbers different?
Manager
Manager
Joined: 14 Aug 2003
Posts: 90
Location: barcelona
Followers: 1

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

 [#permalink] New post 17 Aug 2003, 01:52
mmmmm, lets see...
1. total number of possibilities=6*6*6/3!
number satisfying 2 equal numbers=6*1*5/3!
answer: 5/36
2. total number of possibilities=6^4/4!
number satisfying 0 equal numbers=6*5*4*3/4!
answer: 10/36
3. total number of possibilities=6^6/6!
there is only one way of having different numbers, dont need to calculate that
answer: 1/(6^6/6!)
SVP
SVP
User avatar
Joined: 03 Feb 2003
Posts: 1683
Followers: 4

Kudos [?]: 16 [0], given: 0

 [#permalink] New post 17 Aug 2003, 07:47
my answers:

(1) 5/36
(2) 5/18
(3) 5/324
Intern
Intern
Joined: 18 Jul 2003
Posts: 10
Location: Mumbai
Followers: 0

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

 [#permalink] New post 18 Aug 2003, 02:06
I would like to have the explanation for the 1st one.

I tried the following:-
1) Total no. of possibilities = 6*6*6 = 216
No. of events where all the three are equal = 6
No. of events when none are eqaul = 6*5*4 = 120
Therefore the no. of events when two are equal and third different
= 216-6-120 = 90

Therefore Probability that 2 are equal and third is different = 90 / 216
= 5/12

The remaining two I got the answers same as 5/18 and 5/324 respectively.
Manager
Manager
Joined: 10 Jun 2003
Posts: 216
Location: Maryland
Followers: 2

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

GMAT Tests User
answer [#permalink] New post 18 Aug 2003, 06:25
kheriapiyush@indiatimes.c wrote:
I would like to have the explanation for the 1st one.

I tried the following:-
1) Total no. of possibilities = 6*6*6 = 216
No. of events where all the three are equal = 6
No. of events when none are eqaul = 6*5*4 = 120
Therefore the no. of events when two are equal and third different
= 216-6-120 = 90

Therefore Probability that 2 are equal and third is different = 90 / 216
= 5/12

The remaining two I got the answers same as 5/18 and 5/324 respectively.


I agree with indiatimes on this one. My method was the same:

All three numbers the same we have 6/216.
All three different we have 120/216.
So the remainder is 90/216, reducing to 5/12.

Comments?
Manager
Manager
User avatar
Joined: 03 Jun 2003
Posts: 86
Location: Uruguay
Followers: 1

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

 [#permalink] New post 18 Aug 2003, 06:39
My aproach for the 1st is

first dice can be any number. Second dice must be the same (order does not matter) 1/6. Third dice must be any number but not the same as in 1 and 2, so 5/6

1/6*5/6= 5/36

With number 2 same aproach

5/6*4/6*3/6 = 5/18

With number three same aproach

5/6*4/6*3/6*2/6*1/6 = 5/324
Manager
Manager
Joined: 10 Jun 2003
Posts: 216
Location: Maryland
Followers: 2

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

GMAT Tests User
Answer [#permalink] New post 18 Aug 2003, 07:00
Maybe its just b/c its Monday morning (I hope!!)...

Can sombody please break down there three possibilities of part (A)..

Probability that...

(1) All three are the same
(2) All three are different
(3) Two are the same, and 1 is different
Manager
Manager
User avatar
Joined: 25 Jun 2003
Posts: 100
Followers: 1

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

 [#permalink] New post 20 Aug 2003, 13:25
(1) As explained by Stolyar we have three mutually exclusive situations

(A) S S D ,
(B) S D S ,
(C) D S S

where S = Same number , D = Different Number

p(A) = 1/6 x 1 x 5/6 = 5/36
p(B) = 1/6 x 5/6 x 1 = 5/36
p(C) = 5/6 x 1/6 x 1 = 5/36

Combined Probability = p(A) + p(B) + p(C) = 3 x 5 / 36 = 5/12
_________________

Brainless

SVP
SVP
User avatar
Joined: 03 Feb 2003
Posts: 1683
Followers: 4

Kudos [?]: 16 [0], given: 0

 [#permalink] New post 30 Aug 2003, 23:34
MBA04 wrote:
My aproach for the 1st is

first dice can be any number. Second dice must be the same (order does not matter) 1/6. Third dice must be any number but not the same as in 1 and 2, so 5/6

1/6*5/6= 5/36

With number 2 same aproach

5/6*4/6*3/6 = 5/18

With number three same aproach

5/6*4/6*3/6*2/6*1/6 = 5/324


My approaches are the very same.
SVP
SVP
User avatar
Joined: 03 Feb 2003
Posts: 1683
Followers: 4

Kudos [?]: 16 [0], given: 0

Re: Answer [#permalink] New post 30 Aug 2003, 23:41
mciatto wrote:
Maybe its just b/c its Monday morning (I hope!!)...

Can sombody please break down there three possibilities of part (A)..

Probability that...

(1) All three are the same
(2) All three are different
(3) Two are the same, and 1 is different



(1) 1*1/6*1/6=1/36 (all the three are the same)
(2) 1*5/6*4/6=20/36=5/9 (all the three are different)
(3) 1*1/6*5/6=5/36 (2+1)

Another interesting question:
1/36+20/36+5/36=26/36
the full set of probabilities is 1, so 10/36 is the probability of what?
Re: Answer   [#permalink] 30 Aug 2003, 23:41
    Similar topics Author Replies Last post
Similar
Topics:
 New posts 3 The three fair dice are rolled together. What is the stolyar 4 19 Jun 2003, 01:24
New posts Three fair dice are rolled together. What is the probability stolyar 1 21 Jun 2003, 02:33
New posts three fair dice are rolled together. what is the probability stolyar 7 26 Jun 2003, 00:25
New posts One has three fair dice and rolls them together. What is the Praetorian 1 30 Sep 2003, 17:38
New posts One has four fair dice and rolls them together. What is the Praetorian 3 30 Sep 2003, 17:46
Display posts from previous: Sort by

1) One has three fair dice and rolls them together. What is

  Question banks Downloads My Bookmarks Reviews  
Print view
First unread post

Moderators: Bunuel, VeritasPrepKarishma

Search for:


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®.

Main navigation

GMAT Resources

Partners

MBA Resources

www.gmac.com|www.mba.com