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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

06 Mar 2013, 21:31

1

This post received 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?)?
_________________

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

Answer (D)

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:

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
_________________

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.
_________________

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.
_________________

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

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!!

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.
_________________

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.

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.

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

gmatclubot

Re: If two fair six-sided dice are thrown, what is the probabili
[#permalink]
06 Apr 2015, 00:44

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...