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

 It is currently 26 Jun 2016, 07:40

### 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

# If two fair six-sided dice are thrown, what is the probabili

Author Message
TAGS:

### Hide Tags

Manager
Joined: 28 May 2009
Posts: 155
Location: United States
Concentration: Strategy, General Management
GMAT Date: 03-22-2013
GPA: 3.57
WE: Information Technology (Consulting)
Followers: 6

Kudos [?]: 166 [0], given: 91

If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

06 Mar 2013, 19:21
3
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

64% (02:42) correct 36% (01:37) wrong based on 202 sessions

### HideShow timer Statistics

If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) $$\frac{1}{4}$$

(B) $$\frac{3}{11}$$

(C) $$\frac{5}{18}$$

(D) $$\frac{1}{3}$$

(E) $$\frac{4}{11}$$
[Reveal] Spoiler: OA

_________________

Last edited by Bunuel on 19 Jul 2013, 14:24, edited 1 time in total.
Edited the OA.
Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 559
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 570 Q49 V19
GMAT 2: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Followers: 89

Kudos [?]: 469 [1] , given: 53

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

06 Mar 2013, 21:31
1
KUDOS
megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) $$\frac{1}{4}$$

(B) $$\frac{3}{11}$$

(C) $$\frac{5}{18}$$

(D) $$\frac{1}{3}$$

(E) $$\frac{4}{11}$$

My solution is wrong!
Outcomes are (1,2,3,4,5,6) (1,2,3,4,5,6)
Possible values of sum are
1+1=2, 1+2=3,1+3=4,1+4=5, 1+5=6,1+6=7
2+1,2+2,2+3,2+4,2+5 are already taken into consideration above so not counted again
2+6=8,2+7=9
3+1,3+2,3+3,3+4,3+5,3+6 are already taken into consideration above so not counted again
4+1,4+2,4+3,4+4,4+5 are already taken into consideration above so not counted again
4+6 = 10
5+1,5+2,5+3,5+4,5+5 are already taken into consideration above so not counted again
5+6 = 11
6+1,6+2,6+3,6+4,6+5 are already taken into consideration above so not counted again
6+6 = 12

Total possible values of sum are
2,3,4,5,6,7,8,9,10,11,12
-> Total values = 11
-> number of multiples of 3 are 3,6,9,12 = 4
Probability that sum is a multiple of 3 is 4/11

So, Answer will be E according to me
Am not sure where am going wrong(if i am?)?
_________________

Ankit

Check my Tutoring Site -> Brush My Quant

GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief

How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6667
Location: Pune, India
Followers: 1827

Kudos [?]: 11109 [3] , given: 218

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

06 Mar 2013, 21:36
3
KUDOS
Expert's post
megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) $$\frac{1}{4}$$

(B) $$\frac{3}{11}$$

(C) $$\frac{5}{18}$$

(D) $$\frac{1}{3}$$

(E) $$\frac{4}{11}$$

Method 1: Useful for more than 2 dice too.

Check out these posts first:
http://www.veritasprep.com/blog/2012/10 ... l-picture/
http://www.veritasprep.com/blog/2012/10 ... e-part-ii/

Now, the explanation given below will make sense.

No of ways of obtaining a sum of 2 = No of ways of obtaining a sum of 12
No of ways of obtaining a sum of 3 = No of ways of obtaining a sum of 11
No of ways of obtaining a sum of 4 = No of ways of obtaining a sum of 10
No of ways of obtaining a sum of 5 = No of ways of obtaining a sum of 9
No of ways of obtaining a sum of 6 = No of ways of obtaining a sum of 8

No of ways of obtaining:

A sum of 3: 2C1 = 2
A sum of 6: 5C1 = 5
A sum of 9 = A sum of 5: 4C1 = 4
A sum of 12 = A sum of 2: 1C1 = 1

Total no of ways of obtaining a sum which is a multiple of 3 = 2 + 5 + 4 + 1 = 12

Total no of ways = 6*6 = 36

Probability of the sum being a multiple of 3 = 12/36 = 1/3

Method 2: You could simply count the number of ways of getting 3, 6, 9 and 12

Another Method: Since we have only 2 dice.
Check out this table of the 36 different outcomes:

....1 ......2 ......3 .....4 ......5 ......6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Each column gives you 6 different outcomes which vary by 1 each. Hence there will be exactly 2 outcomes with sum as multiple of 3 (take any 6 consecutive numbers. Exactly 2 of them will be a multiple of 3)

Hence required probability = 1/3
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6667 Location: Pune, India Followers: 1827 Kudos [?]: 11109 [1] , given: 218 Re: If two fair six-sided dice are thrown, what is the probabili [#permalink] ### Show Tags 06 Mar 2013, 21:42 1 This post received KUDOS Expert's post nktdotgupta wrote: Total possible values of sum are 2,3,4,5,6,7,8,9,10,11,12 -> Total values = 11 -> number of multiples of 3 are 3,6,9,12 = 4 Probability that sum is a multiple of 3 is 4/11 So, Answer will be E according to me Am not sure where am going wrong(if i am?)? Where you are going wrong is that the probability of getting a sum of 3 is not the same as the probability of getting 4. You are assuming that each sum is obtained with equal probability. There are 2 ways of obtaining 3 but 3 ways of obtaining 4. Hence, it is more probable that you will get a sum of 4 as compared to a sum of 3. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Math Expert
Joined: 02 Sep 2009
Posts: 33504
Followers: 5931

Kudos [?]: 73507 [0], given: 9902

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

07 Mar 2013, 01:48
Expert's post
megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) $$\frac{1}{4}$$

(B) $$\frac{3}{11}$$

(C) $$\frac{5}{18}$$

(D) $$\frac{1}{3}$$

(E) $$\frac{4}{11}$$

Very good discussion about the same concept: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.
_________________
Intern
Joined: 22 Dec 2012
Posts: 16
GMAT 1: 720 Q49 V39
Followers: 1

Kudos [?]: 20 [0], given: 19

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

09 Mar 2013, 01:40
Hi,

I too got the answer as 1/3 however the OA , points to a different choice.. can someone pls correct/clarify???

Bunuel wrote:
megafan wrote:
If two fair six-sided dice are thrown, what is the probability that the sum of the numbers showing on the dice is a multiple of 3 ?

(A) $$\frac{1}{4}$$

(B) $$\frac{3}{11}$$

(C) $$\frac{5}{18}$$

(D) $$\frac{1}{3}$$

(E) $$\frac{4}{11}$$

Very good discussion about the same concept: mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 10192
Followers: 481

Kudos [?]: 124 [0], given: 0

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

04 Jul 2014, 12:53
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 26 Mar 2015
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 13

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

03 Apr 2015, 03:56
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3
Manager
Joined: 28 Jan 2015
Posts: 132
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38
Followers: 0

Kudos [?]: 19 [0], given: 51

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

05 Apr 2015, 20:37
rohitthakur wrote:
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3

I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6667
Location: Pune, India
Followers: 1827

Kudos [?]: 11109 [0], given: 218

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

05 Apr 2015, 21:02
Expert's post
sabineodf wrote:
rohitthakur wrote:
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3

I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!

Imagine that one die is red and the other is yellow. A 5 on red and 1 on yellow is different from a 1 on red and 5 on yellow.
But a 3 on red and 3 on yellow is the same as 3 on red and 3 on yellow. Hence, you count it only once.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 28 Jan 2015
Posts: 132
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38
Followers: 0

Kudos [?]: 19 [0], given: 51

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

05 Apr 2015, 22:02
VeritasPrepKarishma wrote:
sabineodf wrote:
rohitthakur wrote:
the easy way to do this is,
numbers multiple of 3 are : 3,6,9 and 12, those can appear on twice dice throw,
3 can come on dice in two ways as (1,2) and(2,1)
6 can be there in 5 ways:( 1,5), (5,1), (2,4), (4,2) and (3,3)
9 can be in 4 ways: (4,5), (5,4),(3,6) and (6,3)
now 12 can appear only in one way i.e, (6,6)
adding each probability it has 12 ways so dividing it by 36 total throws will give 1/3

I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!

Imagine that one die is red and the other is yellow. A 5 on red and 1 on yellow is different from a 1 on red and 5 on yellow.
But a 3 on red and 3 on yellow is the same as 3 on red and 3 on yellow. Hence, you count it only once.

Thank you, that was very helpful!
Manager
Joined: 19 Mar 2015
Posts: 81
Followers: 0

Kudos [?]: 9 [0], given: 11

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

06 Apr 2015, 00:37
sabineodf wrote:
I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!

When we say that the total number of combinations possible are 36, it includes combinations such as (1,5),(5,1), (2,3),(3,2),(4,1),(1,4) etc.

So, this itself shows that (5,1) and (1,5) are two separate possibilities.
Manager
Joined: 28 Jan 2015
Posts: 132
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38
Followers: 0

Kudos [?]: 19 [0], given: 51

Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

### Show Tags

06 Apr 2015, 00:44
gmatgrl wrote:
sabineodf wrote:
I had the same approach as you, but I got stuck because I wrote the possibility of getting 6 as one too many, since I put down (3,3) twice. I see that was wrong now, but why is that wrong when (5,1) and (1,5) are two separate possibilities? Thanks if you are able to explain that to me!!

When we say that the total number of combinations possible are 36, it includes combinations such as (1,5),(5,1), (2,3),(3,2),(4,1),(1,4) etc.

So, this itself shows that (5,1) and (1,5) are two separate possibilities.

Yes, I am aware of that. My questions was why rolling (3,3) was not considered to be two separate possibilities... Which was explained to me above
Re: If two fair six-sided dice are thrown, what is the probabili   [#permalink] 06 Apr 2015, 00:44
Similar topics Replies Last post
Similar
Topics:
4 Two fair dice are thrown together. what is the probability that the nu 2 22 Sep 2015, 18:18
5 If four fair dice are thrown simultaneously, what is the probability 4 20 Jun 2015, 03:08
12 Brian plays a game in which two fair, six-sided dice are rol 12 18 Jun 2014, 08:52
3 If two six-sided dice are thrown, what is the probability 6 22 Nov 2012, 15:12
12 A gambler rolls three fair six-sided dice. What is the probability 5 31 Jan 2012, 17:50
Display posts from previous: Sort by