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

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

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06 Mar 2013, 19:21
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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

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Last edited by Bunuel on 19 Jul 2013, 14:24, edited 1 time in total.
Edited the OA.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

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

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06 Mar 2013, 21:36
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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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6830 Location: Pune, India Followers: 1924 Kudos [?]: 11943 [1] , given: 221 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

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

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07 Mar 2013, 01:48
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.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

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

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

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

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

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05 Apr 2015, 21:02
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.
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Re: If two fair six-sided dice are thrown, what is the probabili [#permalink]

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

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

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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
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