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

 It is currently 26 Apr 2015, 02:02

### 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

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

Author Message
TAGS:
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

1) One has three fair dice and rolls them together. What is [#permalink]  17 Aug 2003, 00:38
(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
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

mmmmm, lets see...
1. total number of possibilities=6*6*6/3!
number satisfying 2 equal numbers=6*1*5/3!
2. total number of possibilities=6^4/4!
number satisfying 0 equal numbers=6*5*4*3/4!
3. total number of possibilities=6^6/6!
there is only one way of having different numbers, dont need to calculate that
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

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

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

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
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 4 [0], given: 0

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.

Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay
Followers: 1

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

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
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 4 [0], given: 0

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
Joined: 25 Jun 2003
Posts: 95
Followers: 1

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

(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
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

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
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

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?
Similar topics Replies Last post
Similar
Topics:
4 A gambler rolls 3 fair dice. What is the probability that tw 3 31 Jan 2012, 16:50
2 Gambler: The odds of rolling a three with a fair die is one 11 16 Dec 2010, 09:54
8 Two fair dices are rolled. Find the probability that the 12 16 Nov 2007, 09:55
The six fair dice are rolled together. What is the 11 08 Feb 2006, 10:28
Two fair dices are rolled. Find the probability that the 4 27 Nov 2005, 16:10
Display posts from previous: Sort by